Cálculo Ejemplos

$f (x) = {(2 x)}^{x}$

Paso 1

Obtén la primera derivada de la función.

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Paso 1.1

Simplifica con la obtención del factor común.

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Paso 1.1.1

Factoriza $2$ de $2 x$ .

$\frac{d}{d x} [{(2 (x))}^{x}]$

Paso 1.1.2

Aplica la regla del producto a $2 (x)$ .

$\frac{d}{d x} [2^{x} x^{x}]$

Paso 1.2

Diferencia con la regla del producto, que establece que $\frac{d}{d x} [f (x) g (x)]$ es $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ donde $f (x) = 2^{x}$ y $g (x) = x^{x}$ .

$2^{x} \frac{d}{d x} [x^{x}] + x^{x} \frac{d}{d x} [2^{x}]$

Paso 1.3

Usa las propiedades de los logaritmos para simplificar la diferenciación.

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Paso 1.3.1

Reescribe $x^{x}$ como $e^{\ln (x^{x})}$ .

$2^{x} \frac{d}{d x} [e^{\ln (x^{x})}] + x^{x} \frac{d}{d x} [2^{x}]$

Paso 1.3.2

Expande $\ln (x^{x})$ ; para ello, mueve $x$ fuera del logaritmo.

$2^{x} \frac{d}{d x} [e^{x \ln (x)}] + x^{x} \frac{d}{d x} [2^{x}]$

Paso 1.4

Diferencia con la regla de la cadena, que establece que $\frac{d}{d x} [f (g (x))]$ es $f' (g (x)) g' (x)$ donde $f (x) = e^{x}$ y $g (x) = x \ln (x)$ .

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Paso 1.4.1

Para aplicar la regla de la cadena, establece $u$ como $x \ln (x)$ .

$2^{x} (\frac{d}{d u} [e^{u}] \frac{d}{d x} [x \ln (x)]) + x^{x} \frac{d}{d x} [2^{x}]$

Paso 1.4.2

Diferencia con la regla exponencial, que establece que $\frac{d}{d u} [a^{u}]$ es $a^{u} \ln (a)$ donde $a$ = $e$ .

$2^{x} (e^{u} \frac{d}{d x} [x \ln (x)]) + x^{x} \frac{d}{d x} [2^{x}]$

Paso 1.4.3

Reemplaza todos los casos de $u$ con $x \ln (x)$ .

$2^{x} (e^{x \ln (x)} \frac{d}{d x} [x \ln (x)]) + x^{x} \frac{d}{d x} [2^{x}]$

Paso 1.5

Diferencia con la regla del producto, que establece que $\frac{d}{d x} [f (x) g (x)]$ es $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ donde $f (x) = x$ y $g (x) = \ln (x)$ .

$2^{x} (e^{x \ln (x)} (x \frac{d}{d x} [\ln (x)] + \ln (x) \frac{d}{d x} [x])) + x^{x} \frac{d}{d x} [2^{x}]$

Paso 1.6

La derivada de $\ln (x)$ con respecto a $x$ es $\frac{1}{x}$ .

$2^{x} (e^{x \ln (x)} (x \frac{1}{x} + \ln (x) \frac{d}{d x} [x])) + x^{x} \frac{d}{d x} [2^{x}]$

Paso 1.7

Diferencia con la regla de la potencia.

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Paso 1.7.1

Combina $x$ y $\frac{1}{x}$ .

$2^{x} (e^{x \ln (x)} (\frac{x}{x} + \ln (x) \frac{d}{d x} [x])) + x^{x} \frac{d}{d x} [2^{x}]$

Paso 1.7.2

Cancela el factor común de $x$ .

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Paso 1.7.2.1

Cancela el factor común.

$2^{x} (e^{x \ln (x)} (\frac{x}{x} + \ln (x) \frac{d}{d x} [x])) + x^{x} \frac{d}{d x} [2^{x}]$

Paso 1.7.2.2

Reescribe la expresión.

$2^{x} (e^{x \ln (x)} (1 + \ln (x) \frac{d}{d x} [x])) + x^{x} \frac{d}{d x} [2^{x}]$

Paso 1.7.3

Diferencia con la regla de la potencia, que establece que $\frac{d}{d x} [x^{n}]$ es $n x^{n - 1}$ donde $n = 1$ .

$2^{x} (e^{x \ln (x)} (1 + \ln (x) \cdot 1)) + x^{x} \frac{d}{d x} [2^{x}]$

Paso 1.7.4

Multiplica $\ln (x)$ por $1$ .

$2^{x} (e^{x \ln (x)} (1 + \ln (x))) + x^{x} \frac{d}{d x} [2^{x}]$

Paso 1.8

Diferencia con la regla exponencial, que establece que $\frac{d}{d x} [a^{x}]$ es $a^{x} \ln (a)$ donde $a$ = $2$ .

$2^{x} (e^{x \ln (x)} (1 + \ln (x))) + x^{x} (2^{x} \ln (2))$

Paso 1.9

Simplifica.

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Paso 1.9.1

Aplica la propiedad distributiva.

$2^{x} (e^{x \ln (x)} \cdot 1 + e^{x \ln (x)} \ln (x)) + x^{x} (2^{x} \ln (2))$

Paso 1.9.2

Aplica la propiedad distributiva.

$2^{x} (e^{x \ln (x)} \cdot 1) + 2^{x} (e^{x \ln (x)} \ln (x)) + x^{x} (2^{x} \ln (2))$

Paso 1.9.3

Multiplica $e^{x \ln (x)}$ por $1$ .

$2^{x} e^{x \ln (x)} + 2^{x} (e^{x \ln (x)} \ln (x)) + x^{x} (2^{x} \ln (2))$

Paso 1.9.4

Reordena los términos.

$2^{x} e^{x \ln (x)} + 2^{x} e^{x \ln (x)} \ln (x) + 2^{x} x^{x} \ln (2)$

Paso 2

Obtén la segunda derivada de la función.

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Paso 2.1

Según la regla de la suma, la derivada de $2^{x} e^{x \ln (x)} + 2^{x} e^{x \ln (x)} \ln (x) + 2^{x} x^{x} \ln (2)$ con respecto a $x$ es $\frac{d}{d x} [2^{x} e^{x \ln (x)}] + \frac{d}{d x} [2^{x} e^{x \ln (x)} \ln (x)] + \frac{d}{d x} [2^{x} x^{x} \ln (2)]$ .

$f'' (x) = \frac{d}{d x} (2^{x} e^{x \ln (x)}) + \frac{d}{d x} (2^{x} e^{x \ln (x)} \ln (x)) + \frac{d}{d x} (2^{x} x^{x} \ln (2))$

Paso 2.2

Evalúa $\frac{d}{d x} [2^{x} e^{x \ln (x)}]$ .

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Paso 2.2.1

Diferencia con la regla del producto, que establece que $\frac{d}{d x} [f (x) g (x)]$ es $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ donde $f (x) = 2^{x}$ y $g (x) = e^{x \ln (x)}$ .

$f'' (x) = 2^{x} \frac{d}{d x} (e^{x \ln (x)}) + e^{x \ln (x)} \frac{d}{d x} (2^{x}) + \frac{d}{d x} (2^{x} e^{x \ln (x)} \ln (x)) + \frac{d}{d x} (2^{x} x^{x} \ln (2))$

Paso 2.2.2

Diferencia con la regla de la cadena, que establece que $\frac{d}{d x} [f (g (x))]$ es $f' (g (x)) g' (x)$ donde $f (x) = e^{x}$ y $g (x) = x \ln (x)$ .

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Paso 2.2.2.1

Para aplicar la regla de la cadena, establece $u_{1}$ como $x \ln (x)$ .

$f'' (x) = 2^{x} (\frac{d}{d u (1)} (e^{u_{1}}) \frac{d}{d x} (x \ln (x))) + e^{x \ln (x)} \frac{d}{d x} (2^{x}) + \frac{d}{d x} (2^{x} e^{x \ln (x)} \ln (x)) + \frac{d}{d x} (2^{x} x^{x} \ln (2))$

Paso 2.2.2.2

Diferencia con la regla exponencial, que establece que $\frac{d}{d u_{1}} [a^{u_{1}}]$ es $a^{u_{1}} \ln (a)$ donde $a$ = $e$ .

$f'' (x) = 2^{x} (e^{u_{1}} \frac{d}{d x} (x \ln (x))) + e^{x \ln (x)} \frac{d}{d x} (2^{x}) + \frac{d}{d x} (2^{x} e^{x \ln (x)} \ln (x)) + \frac{d}{d x} (2^{x} x^{x} \ln (2))$

Paso 2.2.2.3

Reemplaza todos los casos de $u_{1}$ con $x \ln (x)$ .

$f'' (x) = 2^{x} (e^{x \ln (x)} \frac{d}{d x} (x \ln (x))) + e^{x \ln (x)} \frac{d}{d x} (2^{x}) + \frac{d}{d x} (2^{x} e^{x \ln (x)} \ln (x)) + \frac{d}{d x} (2^{x} x^{x} \ln (2))$

Paso 2.2.3

Diferencia con la regla del producto, que establece que $\frac{d}{d x} [f (x) g (x)]$ es $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ donde $f (x) = x$ y $g (x) = \ln (x)$ .

$f'' (x) = 2^{x} (e^{x \ln (x)} (x \frac{d}{d x} (\ln (x)) + \ln (x) \frac{d}{d x} (x))) + e^{x \ln (x)} \frac{d}{d x} (2^{x}) + \frac{d}{d x} (2^{x} e^{x \ln (x)} \ln (x)) + \frac{d}{d x} (2^{x} x^{x} \ln (2))$

Paso 2.2.4

La derivada de $\ln (x)$ con respecto a $x$ es $\frac{1}{x}$ .

$f'' (x) = 2^{x} (e^{x \ln (x)} (x (\frac{1}{x}) + \ln (x) \frac{d}{d x} (x))) + e^{x \ln (x)} \frac{d}{d x} (2^{x}) + \frac{d}{d x} (2^{x} e^{x \ln (x)} \ln (x)) + \frac{d}{d x} (2^{x} x^{x} \ln (2))$

Paso 2.2.5

Diferencia con la regla de la potencia, que establece que $\frac{d}{d x} [x^{n}]$ es $n x^{n - 1}$ donde $n = 1$ .

$f'' (x) = 2^{x} (e^{x \ln (x)} (x (\frac{1}{x}) + \ln (x) \cdot 1)) + e^{x \ln (x)} \frac{d}{d x} (2^{x}) + \frac{d}{d x} (2^{x} e^{x \ln (x)} \ln (x)) + \frac{d}{d x} (2^{x} x^{x} \ln (2))$

Paso 2.2.6

Diferencia con la regla exponencial, que establece que $\frac{d}{d x} [a^{x}]$ es $a^{x} \ln (a)$ donde $a$ = $2$ .

$f'' (x) = 2^{x} (e^{x \ln (x)} (x (\frac{1}{x}) + \ln (x) \cdot 1)) + e^{x \ln (x)} (2^{x} \ln (2)) + \frac{d}{d x} (2^{x} e^{x \ln (x)} \ln (x)) + \frac{d}{d x} (2^{x} x^{x} \ln (2))$

Paso 2.2.7

Combina $x$ y $\frac{1}{x}$ .

$f'' (x) = 2^{x} (e^{x \ln (x)} (\frac{x}{x} + \ln (x) \cdot 1)) + e^{x \ln (x)} (2^{x} \ln (2)) + \frac{d}{d x} (2^{x} e^{x \ln (x)} \ln (x)) + \frac{d}{d x} (2^{x} x^{x} \ln (2))$

Paso 2.2.8

Cancela el factor común de $x$ .

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Paso 2.2.8.1

Cancela el factor común.

$f'' (x) = 2^{x} (e^{x \ln (x)} (\frac{x}{x} + \ln (x) \cdot 1)) + e^{x \ln (x)} (2^{x} \ln (2)) + \frac{d}{d x} (2^{x} e^{x \ln (x)} \ln (x)) + \frac{d}{d x} (2^{x} x^{x} \ln (2))$

Paso 2.2.8.2

Reescribe la expresión.

$f'' (x) = 2^{x} (e^{x \ln (x)} (1 + \ln (x) \cdot 1)) + e^{x \ln (x)} (2^{x} \ln (2)) + \frac{d}{d x} (2^{x} e^{x \ln (x)} \ln (x)) + \frac{d}{d x} (2^{x} x^{x} \ln (2))$

Paso 2.2.9

Multiplica $\ln (x)$ por $1$ .

$f'' (x) = 2^{x} (e^{x \ln (x)} (1 + \ln (x))) + e^{x \ln (x)} (2^{x} \ln (2)) + \frac{d}{d x} (2^{x} e^{x \ln (x)} \ln (x)) + \frac{d}{d x} (2^{x} x^{x} \ln (2))$

Paso 2.3

Evalúa $\frac{d}{d x} [2^{x} e^{x \ln (x)} \ln (x)]$ .

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Paso 2.3.1

Diferencia con la regla del producto, que establece que $\frac{d}{d x} [f (x) g (x)]$ es $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ donde $f (x) = 2^{x} e^{x \ln (x)}$ y $g (x) = \ln (x)$ .

$f'' (x) = 2^{x} (e^{x \ln (x)} (1 + \ln (x))) + e^{x \ln (x)} (2^{x} \ln (2)) + 2^{x} e^{x \ln (x)} \frac{d}{d x} (\ln (x)) + \ln (x) \frac{d}{d x} (2^{x} e^{x \ln (x)}) + \frac{d}{d x} (2^{x} x^{x} \ln (2))$

Paso 2.3.2

La derivada de $\ln (x)$ con respecto a $x$ es $\frac{1}{x}$ .

$f'' (x) = 2^{x} (e^{x \ln (x)} (1 + \ln (x))) + e^{x \ln (x)} (2^{x} \ln (2)) + 2^{x} e^{x \ln (x)} (\frac{1}{x}) + \ln (x) \frac{d}{d x} (2^{x} e^{x \ln (x)}) + \frac{d}{d x} (2^{x} x^{x} \ln (2))$

Paso 2.3.3

Diferencia con la regla del producto, que establece que $\frac{d}{d x} [f (x) g (x)]$ es $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ donde $f (x) = 2^{x}$ y $g (x) = e^{x \ln (x)}$ .

$f'' (x) = 2^{x} (e^{x \ln (x)} (1 + \ln (x))) + e^{x \ln (x)} (2^{x} \ln (2)) + 2^{x} e^{x \ln (x)} (\frac{1}{x}) + \ln (x) (2^{x} \frac{d}{d x} (e^{x \ln (x)}) + e^{x \ln (x)} \frac{d}{d x} (2^{x})) + \frac{d}{d x} (2^{x} x^{x} \ln (2))$

Paso 2.3.4

Diferencia con la regla de la cadena, que establece que $\frac{d}{d x} [f (g (x))]$ es $f' (g (x)) g' (x)$ donde $f (x) = e^{x}$ y $g (x) = x \ln (x)$ .

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Paso 2.3.4.1

Para aplicar la regla de la cadena, establece $u_{2}$ como $x \ln (x)$ .

$f'' (x) = 2^{x} (e^{x \ln (x)} (1 + \ln (x))) + e^{x \ln (x)} (2^{x} \ln (2)) + 2^{x} e^{x \ln (x)} (\frac{1}{x}) + \ln (x) (2^{x} (\frac{d}{d u (2)} (e^{u_{2}}) \frac{d}{d x} (x \ln (x))) + e^{x \ln (x)} \frac{d}{d x} (2^{x})) + \frac{d}{d x} (2^{x} x^{x} \ln (2))$

Paso 2.3.4.2

Diferencia con la regla exponencial, que establece que $\frac{d}{d u_{2}} [a^{u_{2}}]$ es $a^{u_{2}} \ln (a)$ donde $a$ = $e$ .

$f'' (x) = 2^{x} (e^{x \ln (x)} (1 + \ln (x))) + e^{x \ln (x)} (2^{x} \ln (2)) + 2^{x} e^{x \ln (x)} (\frac{1}{x}) + \ln (x) (2^{x} (e^{u_{2}} \frac{d}{d x} (x \ln (x))) + e^{x \ln (x)} \frac{d}{d x} (2^{x})) + \frac{d}{d x} (2^{x} x^{x} \ln (2))$

Paso 2.3.4.3

Reemplaza todos los casos de $u_{2}$ con $x \ln (x)$ .

$f'' (x) = 2^{x} (e^{x \ln (x)} (1 + \ln (x))) + e^{x \ln (x)} (2^{x} \ln (2)) + 2^{x} e^{x \ln (x)} (\frac{1}{x}) + \ln (x) (2^{x} (e^{x \ln (x)} \frac{d}{d x} (x \ln (x))) + e^{x \ln (x)} \frac{d}{d x} (2^{x})) + \frac{d}{d x} (2^{x} x^{x} \ln (2))$

Paso 2.3.5

Diferencia con la regla del producto, que establece que $\frac{d}{d x} [f (x) g (x)]$ es $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ donde $f (x) = x$ y $g (x) = \ln (x)$ .

$f'' (x) = 2^{x} (e^{x \ln (x)} (1 + \ln (x))) + e^{x \ln (x)} (2^{x} \ln (2)) + 2^{x} e^{x \ln (x)} (\frac{1}{x}) + \ln (x) (2^{x} (e^{x \ln (x)} (x \frac{d}{d x} (\ln (x)) + \ln (x) \frac{d}{d x} (x))) + e^{x \ln (x)} \frac{d}{d x} (2^{x})) + \frac{d}{d x} (2^{x} x^{x} \ln (2))$

Paso 2.3.6

La derivada de $\ln (x)$ con respecto a $x$ es $\frac{1}{x}$ .

$f'' (x) = 2^{x} (e^{x \ln (x)} (1 + \ln (x))) + e^{x \ln (x)} (2^{x} \ln (2)) + 2^{x} e^{x \ln (x)} (\frac{1}{x}) + \ln (x) (2^{x} (e^{x \ln (x)} (x (\frac{1}{x}) + \ln (x) \frac{d}{d x} (x))) + e^{x \ln (x)} \frac{d}{d x} (2^{x})) + \frac{d}{d x} (2^{x} x^{x} \ln (2))$

Paso 2.3.7

Diferencia con la regla de la potencia, que establece que $\frac{d}{d x} [x^{n}]$ es $n x^{n - 1}$ donde $n = 1$ .

$f'' (x) = 2^{x} (e^{x \ln (x)} (1 + \ln (x))) + e^{x \ln (x)} (2^{x} \ln (2)) + 2^{x} e^{x \ln (x)} (\frac{1}{x}) + \ln (x) (2^{x} (e^{x \ln (x)} (x (\frac{1}{x}) + \ln (x) \cdot 1)) + e^{x \ln (x)} \frac{d}{d x} (2^{x})) + \frac{d}{d x} (2^{x} x^{x} \ln (2))$

Paso 2.3.8

Diferencia con la regla exponencial, que establece que $\frac{d}{d x} [a^{x}]$ es $a^{x} \ln (a)$ donde $a$ = $2$ .

$f'' (x) = 2^{x} (e^{x \ln (x)} (1 + \ln (x))) + e^{x \ln (x)} (2^{x} \ln (2)) + 2^{x} e^{x \ln (x)} (\frac{1}{x}) + \ln (x) (2^{x} (e^{x \ln (x)} (x (\frac{1}{x}) + \ln (x) \cdot 1)) + e^{x \ln (x)} (2^{x} \ln (2))) + \frac{d}{d x} (2^{x} x^{x} \ln (2))$

Paso 2.3.9

Combina $\frac{1}{x}$ y $2^{x}$ .

$f'' (x) = 2^{x} (e^{x \ln (x)} (1 + \ln (x))) + e^{x \ln (x)} (2^{x} \ln (2)) + \frac{2^{x}}{x} \cdot e^{x \ln (x)} + \ln (x) (2^{x} (e^{x \ln (x)} (x (\frac{1}{x}) + \ln (x) \cdot 1)) + e^{x \ln (x)} (2^{x} \ln (2))) + \frac{d}{d x} (2^{x} x^{x} \ln (2))$

Paso 2.3.10

Combina $\frac{2^{x}}{x}$ y $e^{x \ln (x)}$ .

$f'' (x) = 2^{x} (e^{x \ln (x)} (1 + \ln (x))) + e^{x \ln (x)} (2^{x} \ln (2)) + \frac{2^{x} e^{x \ln (x)}}{x} + \ln (x) (2^{x} (e^{x \ln (x)} (x (\frac{1}{x}) + \ln (x) \cdot 1)) + e^{x \ln (x)} (2^{x} \ln (2))) + \frac{d}{d x} (2^{x} x^{x} \ln (2))$

Paso 2.3.11

Combina $x$ y $\frac{1}{x}$ .

$f'' (x) = 2^{x} (e^{x \ln (x)} (1 + \ln (x))) + e^{x \ln (x)} (2^{x} \ln (2)) + \frac{2^{x} e^{x \ln (x)}}{x} + \ln (x) (2^{x} (e^{x \ln (x)} (\frac{x}{x} + \ln (x) \cdot 1)) + e^{x \ln (x)} (2^{x} \ln (2))) + \frac{d}{d x} (2^{x} x^{x} \ln (2))$

Paso 2.3.12

Cancela el factor común de $x$ .

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Paso 2.3.12.1

Cancela el factor común.

Paso 2.3.12.2

Reescribe la expresión.

$f'' (x) = 2^{x} (e^{x \ln (x)} (1 + \ln (x))) + e^{x \ln (x)} (2^{x} \ln (2)) + \frac{2^{x} e^{x \ln (x)}}{x} + \ln (x) (2^{x} (e^{x \ln (x)} (1 + \ln (x) \cdot 1)) + e^{x \ln (x)} (2^{x} \ln (2))) + \frac{d}{d x} (2^{x} x^{x} \ln (2))$

Paso 2.3.13

Multiplica $\ln (x)$ por $1$ .

$f'' (x) = 2^{x} (e^{x \ln (x)} (1 + \ln (x))) + e^{x \ln (x)} (2^{x} \ln (2)) + \frac{2^{x} e^{x \ln (x)}}{x} + \ln (x) (2^{x} (e^{x \ln (x)} (1 + \ln (x))) + e^{x \ln (x)} (2^{x} \ln (2))) + \frac{d}{d x} (2^{x} x^{x} \ln (2))$

Paso 2.3.14

Reordena los factores de $\ln (x) (2^{x} (e^{x \ln (x)} (1 + \ln (x))) + e^{x \ln (x)} (2^{x} \ln (2)))$ .

$f'' (x) = 2^{x} (e^{x \ln (x)} (1 + \ln (x))) + e^{x \ln (x)} (2^{x} \ln (2)) + \frac{2^{x} e^{x \ln (x)}}{x} + (2^{x} (e^{x \ln (x)} (1 + \ln (x))) + e^{x \ln (x)} (2^{x} \ln (2))) \ln (x) + \frac{d}{d x} (2^{x} x^{x} \ln (2))$

Paso 2.3.15

Para escribir $(2^{x} (e^{x \ln (x)} (1 + \ln (x))) + e^{x \ln (x)} (2^{x} \ln (2))) \ln (x)$ como una fracción con un denominador común, multiplica por $\frac{x}{x}$ .

$f'' (x) = 2^{x} (e^{x \ln (x)} (1 + \ln (x))) + e^{x \ln (x)} (2^{x} \ln (2)) + \frac{2^{x} e^{x \ln (x)}}{x} + \frac{(2^{x} (e^{x \ln (x)} (1 + \ln (x))) + e^{x \ln (x)} (2^{x} \ln (2))) \ln (x) x}{x} + \frac{d}{d x} (2^{x} x^{x} \ln (2))$

Paso 2.3.16

Combina los numeradores sobre el denominador común.

$f'' (x) = 2^{x} (e^{x \ln (x)} (1 + \ln (x))) + e^{x \ln (x)} (2^{x} \ln (2)) + \frac{2^{x} e^{x \ln (x)} + (2^{x} (e^{x \ln (x)} (1 + \ln (x))) + e^{x \ln (x)} (2^{x} \ln (2))) \ln (x) x}{x} + \frac{d}{d x} (2^{x} x^{x} \ln (2))$

Paso 2.4

Evalúa $\frac{d}{d x} [2^{x} x^{x} \ln (2)]$ .

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Paso 2.4.1

Como $\ln (2)$ es constante con respecto a $x$ , la derivada de $2^{x} x^{x} \ln (2)$ con respecto a $x$ es $\ln (2) \frac{d}{d x} [2^{x} x^{x}]$ .

Paso 2.4.2

Diferencia con la regla del producto, que establece que $\frac{d}{d x} [f (x) g (x)]$ es $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ donde $f (x) = 2^{x}$ y $g (x) = x^{x}$ .

Paso 2.4.3

Usa las propiedades de los logaritmos para simplificar la diferenciación.

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Paso 2.4.3.1

Reescribe $x^{x}$ como $e^{\ln (x^{x})}$ .

Paso 2.4.3.2

Expande $\ln (x^{x})$ ; para ello, mueve $x$ fuera del logaritmo.

Paso 2.4.4

Diferencia con la regla de la cadena, que establece que $\frac{d}{d x} [f (g (x))]$ es $f' (g (x)) g' (x)$ donde $f (x) = e^{x}$ y $g (x) = x \ln (x)$ .

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Paso 2.4.4.1

Para aplicar la regla de la cadena, establece $u_{3}$ como $x \ln (x)$ .

Paso 2.4.4.2

Diferencia con la regla exponencial, que establece que $\frac{d}{d u_{3}} [a^{u_{3}}]$ es $a^{u_{3}} \ln (a)$ donde $a$ = $e$ .

Paso 2.4.4.3

Reemplaza todos los casos de $u_{3}$ con $x \ln (x)$ .

Paso 2.4.5

Diferencia con la regla del producto, que establece que $\frac{d}{d x} [f (x) g (x)]$ es $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ donde $f (x) = x$ y $g (x) = \ln (x)$ .

Paso 2.4.6

La derivada de $\ln (x)$ con respecto a $x$ es $\frac{1}{x}$ .

Paso 2.4.7

Diferencia con la regla de la potencia, que establece que $\frac{d}{d x} [x^{n}]$ es $n x^{n - 1}$ donde $n = 1$ .

Paso 2.4.8

Diferencia con la regla exponencial, que establece que $\frac{d}{d x} [a^{x}]$ es $a^{x} \ln (a)$ donde $a$ = $2$ .

Paso 2.4.9

Combina $x$ y $\frac{1}{x}$ .

Paso 2.4.10

Cancela el factor común de $x$ .

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Paso 2.4.10.1

Cancela el factor común.

Paso 2.4.10.2

Reescribe la expresión.

Paso 2.4.11

Multiplica $\ln (x)$ por $1$ .

Paso 2.5

Simplifica.

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Paso 2.5.1

Aplica la propiedad distributiva.

$f'' (x) = 2^{x} (e^{x \ln (x)} \cdot 1 + e^{x \ln (x)} \ln (x)) + e^{x \ln (x)} (2^{x} \ln (2)) + \frac{2^{x} e^{x \ln (x)} + (2^{x} (e^{x \ln (x)} (1 + \ln (x))) + e^{x \ln (x)} (2^{x} \ln (2))) \ln (x) x}{x} + \ln (2) (2^{x} (e^{x \ln (x)} (1 + \ln (x))) + x^{x} (2^{x} \ln (2)))$

Paso 2.5.2

Aplica la propiedad distributiva.

$f'' (x) = 2^{x} (e^{x \ln (x)} \cdot 1) + 2^{x} (e^{x \ln (x)} \ln (x)) + e^{x \ln (x)} (2^{x} \ln (2)) + \frac{2^{x} e^{x \ln (x)} + (2^{x} (e^{x \ln (x)} (1 + \ln (x))) + e^{x \ln (x)} (2^{x} \ln (2))) \ln (x) x}{x} + \ln (2) (2^{x} (e^{x \ln (x)} (1 + \ln (x))) + x^{x} (2^{x} \ln (2)))$

Paso 2.5.3

Aplica la propiedad distributiva.

$f'' (x) = 2^{x} (e^{x \ln (x)} \cdot 1) + 2^{x} (e^{x \ln (x)} \ln (x)) + e^{x \ln (x)} (2^{x} \ln (2)) + \frac{2^{x} e^{x \ln (x)} + (2^{x} (e^{x \ln (x)} \cdot 1 + e^{x \ln (x)} \ln (x)) + e^{x \ln (x)} (2^{x} \ln (2))) \ln (x) x}{x} + \ln (2) (2^{x} (e^{x \ln (x)} (1 + \ln (x))) + x^{x} (2^{x} \ln (2)))$

Paso 2.5.4

Aplica la propiedad distributiva.

$f'' (x) = 2^{x} (e^{x \ln (x)} \cdot 1) + 2^{x} (e^{x \ln (x)} \ln (x)) + e^{x \ln (x)} (2^{x} \ln (2)) + \frac{2^{x} e^{x \ln (x)} + (2^{x} (e^{x \ln (x)} \cdot 1) + 2^{x} (e^{x \ln (x)} \ln (x)) + e^{x \ln (x)} (2^{x} \ln (2))) \ln (x) x}{x} + \ln (2) (2^{x} (e^{x \ln (x)} (1 + \ln (x))) + x^{x} (2^{x} \ln (2)))$

Paso 2.5.5

Aplica la propiedad distributiva.

$f'' (x) = 2^{x} (e^{x \ln (x)} \cdot 1) + 2^{x} (e^{x \ln (x)} \ln (x)) + e^{x \ln (x)} (2^{x} \ln (2)) + \frac{2^{x} e^{x \ln (x)} + (2^{x} (e^{x \ln (x)} \cdot 1) \ln (x) + 2^{x} (e^{x \ln (x)} \ln (x)) \ln (x) + e^{x \ln (x)} (2^{x} \ln (2)) \ln (x)) x}{x} + \ln (2) (2^{x} (e^{x \ln (x)} (1 + \ln (x))) + x^{x} (2^{x} \ln (2)))$

Paso 2.5.6

Aplica la propiedad distributiva.

$f'' (x) = 2^{x} (e^{x \ln (x)} \cdot 1) + 2^{x} (e^{x \ln (x)} \ln (x)) + e^{x \ln (x)} (2^{x} \ln (2)) + \frac{2^{x} e^{x \ln (x)} + 2^{x} (e^{x \ln (x)} \cdot 1) \ln (x) x + 2^{x} (e^{x \ln (x)} \ln (x)) \ln (x) x + e^{x \ln (x)} (2^{x} \ln (2)) \ln (x) x}{x} + \ln (2) (2^{x} (e^{x \ln (x)} (1 + \ln (x))) + x^{x} (2^{x} \ln (2)))$

Paso 2.5.7

Aplica la propiedad distributiva.

Paso 2.5.8

Aplica la propiedad distributiva.

Paso 2.5.9

Aplica la propiedad distributiva.

Paso 2.5.10

Combina los términos.

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Paso 2.5.10.1

Multiplica $e^{x \ln (x)}$ por $1$ .

$f'' (x) = 2^{x} e^{x \ln (x)} + 2^{x} (e^{x \ln (x)} \ln (x)) + e^{x \ln (x)} (2^{x} \ln (2)) + \frac{2^{x} e^{x \ln (x)} + 2^{x} (e^{x \ln (x)} \cdot 1) \ln (x) x + 2^{x} (e^{x \ln (x)} \ln (x)) \ln (x) x + e^{x \ln (x)} (2^{x} \ln (2)) \ln (x) x}{x} + \ln (2) (2^{x} (e^{x \ln (x)} \cdot 1)) + \ln (2) (2^{x} (e^{x \ln (x)} \ln (x))) + \ln (2) (x^{x} (2^{x} \ln (2)))$

Paso 2.5.10.2

Multiplica $e^{x \ln (x)}$ por $1$ .

$f'' (x) = 2^{x} e^{x \ln (x)} + 2^{x} (e^{x \ln (x)} \ln (x)) + e^{x \ln (x)} (2^{x} \ln (2)) + \frac{2^{x} e^{x \ln (x)} + 2^{x} e^{x \ln (x)} \ln (x) x + 2^{x} (e^{x \ln (x)} \ln (x)) \ln (x) x + e^{x \ln (x)} (2^{x} \ln (2)) \ln (x) x}{x} + \ln (2) (2^{x} (e^{x \ln (x)} \cdot 1)) + \ln (2) (2^{x} (e^{x \ln (x)} \ln (x))) + \ln (2) (x^{x} (2^{x} \ln (2)))$

Paso 2.5.10.3

Eleva $\ln (x)$ a la potencia de $1$ .

$f'' (x) = 2^{x} e^{x \ln (x)} + 2^{x} (e^{x \ln (x)} \ln (x)) + e^{x \ln (x)} (2^{x} \ln (2)) + \frac{2^{x} e^{x \ln (x)} + 2^{x} e^{x \ln (x)} \ln (x) x + 2^{x} (e^{x \ln (x)} ((\ln (x)) \ln (x))) x + e^{x \ln (x)} (2^{x} \ln (2)) \ln (x) x}{x} + \ln (2) (2^{x} (e^{x \ln (x)} \cdot 1)) + \ln (2) (2^{x} (e^{x \ln (x)} \ln (x))) + \ln (2) (x^{x} (2^{x} \ln (2)))$

Paso 2.5.10.4

Eleva $\ln (x)$ a la potencia de $1$ .

$f'' (x) = 2^{x} e^{x \ln (x)} + 2^{x} (e^{x \ln (x)} \ln (x)) + e^{x \ln (x)} (2^{x} \ln (2)) + \frac{2^{x} e^{x \ln (x)} + 2^{x} e^{x \ln (x)} \ln (x) x + 2^{x} (e^{x \ln (x)} ((\ln (x)) (\ln (x)))) x + e^{x \ln (x)} (2^{x} \ln (2)) \ln (x) x}{x} + \ln (2) (2^{x} (e^{x \ln (x)} \cdot 1)) + \ln (2) (2^{x} (e^{x \ln (x)} \ln (x))) + \ln (2) (x^{x} (2^{x} \ln (2)))$

Paso 2.5.10.5

Usa la regla de la potencia $a^{m} a^{n} = a^{m + n}$ para combinar exponentes.

$f'' (x) = 2^{x} e^{x \ln (x)} + 2^{x} (e^{x \ln (x)} \ln (x)) + e^{x \ln (x)} (2^{x} \ln (2)) + \frac{2^{x} e^{x \ln (x)} + 2^{x} e^{x \ln (x)} \ln (x) x + 2^{x} (e^{x \ln (x)} {(\ln (x))}^{1 + 1}) x + e^{x \ln (x)} (2^{x} \ln (2)) \ln (x) x}{x} + \ln (2) (2^{x} (e^{x \ln (x)} \cdot 1)) + \ln (2) (2^{x} (e^{x \ln (x)} \ln (x))) + \ln (2) (x^{x} (2^{x} \ln (2)))$

Paso 2.5.10.6

Suma $1$ y $1$ .

$f'' (x) = 2^{x} e^{x \ln (x)} + 2^{x} (e^{x \ln (x)} \ln (x)) + e^{x \ln (x)} (2^{x} \ln (2)) + \frac{2^{x} e^{x \ln (x)} + 2^{x} e^{x \ln (x)} \ln (x) x + 2^{x} (e^{x \ln (x)} \ln^{2} (x)) x + e^{x \ln (x)} (2^{x} \ln (2)) \ln (x) x}{x} + \ln (2) (2^{x} (e^{x \ln (x)} \cdot 1)) + \ln (2) (2^{x} (e^{x \ln (x)} \ln (x))) + \ln (2) (x^{x} (2^{x} \ln (2)))$

Paso 2.5.10.7

Para escribir $2^{x} e^{x \ln (x)}$ como una fracción con un denominador común, multiplica por $\frac{x}{x}$ .

$f'' (x) = 2^{x} (e^{x \ln (x)} \ln (x)) + e^{x \ln (x)} (2^{x} \ln (2)) + \frac{2^{x} e^{x \ln (x)} x}{x} + \frac{2^{x} e^{x \ln (x)} + 2^{x} e^{x \ln (x)} \ln (x) x + 2^{x} (e^{x \ln (x)} \ln^{2} (x)) x + e^{x \ln (x)} (2^{x} \ln (2)) \ln (x) x}{x} + \ln (2) (2^{x} (e^{x \ln (x)} \cdot 1)) + \ln (2) (2^{x} (e^{x \ln (x)} \ln (x))) + \ln (2) (x^{x} (2^{x} \ln (2)))$

Paso 2.5.10.8

Combina los numeradores sobre el denominador común.

$f'' (x) = 2^{x} (e^{x \ln (x)} \ln (x)) + e^{x \ln (x)} (2^{x} \ln (2)) + \frac{2^{x} e^{x \ln (x)} x + 2^{x} e^{x \ln (x)} + 2^{x} e^{x \ln (x)} \ln (x) x + 2^{x} (e^{x \ln (x)} \ln^{2} (x)) x + e^{x \ln (x)} (2^{x} \ln (2)) \ln (x) x}{x} + \ln (2) (2^{x} (e^{x \ln (x)} \cdot 1)) + \ln (2) (2^{x} (e^{x \ln (x)} \ln (x))) + \ln (2) (x^{x} (2^{x} \ln (2)))$

Paso 2.5.10.9

Para escribir $2^{x} (e^{x \ln (x)} \ln (x))$ como una fracción con un denominador común, multiplica por $\frac{x}{x}$ .

$f'' (x) = e^{x \ln (x)} (2^{x} \ln (2)) + \frac{2^{x} (e^{x \ln (x)} \ln (x)) x}{x} + \frac{2^{x} e^{x \ln (x)} x + 2^{x} e^{x \ln (x)} + 2^{x} e^{x \ln (x)} \ln (x) x + 2^{x} (e^{x \ln (x)} \ln^{2} (x)) x + e^{x \ln (x)} (2^{x} \ln (2)) \ln (x) x}{x} + \ln (2) (2^{x} (e^{x \ln (x)} \cdot 1)) + \ln (2) (2^{x} (e^{x \ln (x)} \ln (x))) + \ln (2) (x^{x} (2^{x} \ln (2)))$

Paso 2.5.10.10

Combina los numeradores sobre el denominador común.

$f'' (x) = e^{x \ln (x)} (2^{x} \ln (2)) + \frac{2^{x} (e^{x \ln (x)} \ln (x)) x + 2^{x} e^{x \ln (x)} x + 2^{x} e^{x \ln (x)} + 2^{x} e^{x \ln (x)} \ln (x) x + 2^{x} (e^{x \ln (x)} \ln^{2} (x)) x + e^{x \ln (x)} (2^{x} \ln (2)) \ln (x) x}{x} + \ln (2) (2^{x} (e^{x \ln (x)} \cdot 1)) + \ln (2) (2^{x} (e^{x \ln (x)} \ln (x))) + \ln (2) (x^{x} (2^{x} \ln (2)))$

Paso 2.5.10.11

Mueve $e^{x \ln (x)}$ .

$f'' (x) = e^{x \ln (x)} (2^{x} \ln (2)) + \frac{2^{x} x e^{x \ln (x)} \ln (x) + 2^{x} x e^{x \ln (x)} \ln (x) + 2^{x} e^{x \ln (x)} x + 2^{x} e^{x \ln (x)} + 2^{x} (e^{x \ln (x)} \ln^{2} (x)) x + e^{x \ln (x)} (2^{x} \ln (2)) \ln (x) x}{x} + \ln (2) (2^{x} (e^{x \ln (x)} \cdot 1)) + \ln (2) (2^{x} (e^{x \ln (x)} \ln (x))) + \ln (2) (x^{x} (2^{x} \ln (2)))$

Paso 2.5.10.12

Suma $2^{x} x e^{x \ln (x)} \ln (x)$ y $2^{x} x e^{x \ln (x)} \ln (x)$ .

$f'' (x) = e^{x \ln (x)} (2^{x} \ln (2)) + \frac{2 \cdot (2^{x} x e^{x \ln (x)} \ln (x)) + 2^{x} e^{x \ln (x)} x + 2^{x} e^{x \ln (x)} + 2^{x} (e^{x \ln (x)} \ln^{2} (x)) x + e^{x \ln (x)} (2^{x} \ln (2)) \ln (x) x}{x} + \ln (2) (2^{x} (e^{x \ln (x)} \cdot 1)) + \ln (2) (2^{x} (e^{x \ln (x)} \ln (x))) + \ln (2) (x^{x} (2^{x} \ln (2)))$

Paso 2.5.10.13

Multiplica $2$ por $2^{x}$ .

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Paso 2.5.10.13.1

Eleva $2$ a la potencia de $1$ .

Paso 2.5.10.13.2

Usa la regla de la potencia $a^{m} a^{n} = a^{m + n}$ para combinar exponentes.

$f'' (x) = e^{x \ln (x)} (2^{x} \ln (2)) + \frac{2^{1 + x} x e^{x \ln (x)} \ln (x) + 2^{x} e^{x \ln (x)} x + 2^{x} e^{x \ln (x)} + 2^{x} (e^{x \ln (x)} \ln^{2} (x)) x + e^{x \ln (x)} (2^{x} \ln (2)) \ln (x) x}{x} + \ln (2) (2^{x} (e^{x \ln (x)} \cdot 1)) + \ln (2) (2^{x} (e^{x \ln (x)} \ln (x))) + \ln (2) (x^{x} (2^{x} \ln (2)))$

Paso 2.5.10.14

Para escribir $e^{x \ln (x)} (2^{x} \ln (2))$ como una fracción con un denominador común, multiplica por $\frac{x}{x}$ .

$f'' (x) = \frac{e^{x \ln (x)} (2^{x} \ln (2)) x}{x} + \frac{2^{1 + x} x e^{x \ln (x)} \ln (x) + 2^{x} e^{x \ln (x)} x + 2^{x} e^{x \ln (x)} + 2^{x} (e^{x \ln (x)} \ln^{2} (x)) x + e^{x \ln (x)} (2^{x} \ln (2)) \ln (x) x}{x} + \ln (2) (2^{x} (e^{x \ln (x)} \cdot 1)) + \ln (2) (2^{x} (e^{x \ln (x)} \ln (x))) + \ln (2) (x^{x} (2^{x} \ln (2)))$

Paso 2.5.10.15

Combina los numeradores sobre el denominador común.

$f'' (x) = \frac{e^{x \ln (x)} (2^{x} \ln (2)) x + 2^{1 + x} x e^{x \ln (x)} \ln (x) + 2^{x} e^{x \ln (x)} x + 2^{x} e^{x \ln (x)} + 2^{x} (e^{x \ln (x)} \ln^{2} (x)) x + e^{x \ln (x)} (2^{x} \ln (2)) \ln (x) x}{x} + \ln (2) (2^{x} (e^{x \ln (x)} \cdot 1)) + \ln (2) (2^{x} (e^{x \ln (x)} \ln (x))) + \ln (2) (x^{x} (2^{x} \ln (2)))$

Paso 2.5.10.16

Multiplica $e^{x \ln (x)}$ por $1$ .

$f'' (x) = \frac{e^{x \ln (x)} (2^{x} \ln (2)) x + 2^{1 + x} x e^{x \ln (x)} \ln (x) + 2^{x} e^{x \ln (x)} x + 2^{x} e^{x \ln (x)} + 2^{x} (e^{x \ln (x)} \ln^{2} (x)) x + e^{x \ln (x)} (2^{x} \ln (2)) \ln (x) x}{x} + \ln (2) (2^{x} e^{x \ln (x)}) + \ln (2) (2^{x} (e^{x \ln (x)} \ln (x))) + \ln (2) (x^{x} (2^{x} \ln (2)))$

Paso 2.5.10.17

Eleva $\ln (2)$ a la potencia de $1$ .

$f'' (x) = \frac{e^{x \ln (x)} (2^{x} \ln (2)) x + 2^{1 + x} x e^{x \ln (x)} \ln (x) + 2^{x} e^{x \ln (x)} x + 2^{x} e^{x \ln (x)} + 2^{x} (e^{x \ln (x)} \ln^{2} (x)) x + e^{x \ln (x)} (2^{x} \ln (2)) \ln (x) x}{x} + \ln (2) \cdot (2^{x} e^{x \ln (x)}) + \ln (2) \cdot (2^{x} (e^{x \ln (x)} \ln (x))) + x^{x} (2^{x} ((\ln (2)) \ln (2)))$

Paso 2.5.10.18

Eleva $\ln (2)$ a la potencia de $1$ .

$f'' (x) = \frac{e^{x \ln (x)} (2^{x} \ln (2)) x + 2^{1 + x} x e^{x \ln (x)} \ln (x) + 2^{x} e^{x \ln (x)} x + 2^{x} e^{x \ln (x)} + 2^{x} (e^{x \ln (x)} \ln^{2} (x)) x + e^{x \ln (x)} (2^{x} \ln (2)) \ln (x) x}{x} + \ln (2) \cdot (2^{x} e^{x \ln (x)}) + \ln (2) \cdot (2^{x} (e^{x \ln (x)} \ln (x))) + x^{x} (2^{x} ((\ln (2)) (\ln (2))))$

Paso 2.5.10.19

Usa la regla de la potencia $a^{m} a^{n} = a^{m + n}$ para combinar exponentes.

$f'' (x) = \frac{e^{x \ln (x)} (2^{x} \ln (2)) x + 2^{1 + x} x e^{x \ln (x)} \ln (x) + 2^{x} e^{x \ln (x)} x + 2^{x} e^{x \ln (x)} + 2^{x} (e^{x \ln (x)} \ln^{2} (x)) x + e^{x \ln (x)} (2^{x} \ln (2)) \ln (x) x}{x} + \ln (2) \cdot (2^{x} e^{x \ln (x)}) + \ln (2) \cdot (2^{x} (e^{x \ln (x)} \ln (x))) + x^{x} (2^{x} {(\ln (2))}^{1 + 1})$

Paso 2.5.10.20

Suma $1$ y $1$ .

$f'' (x) = \frac{e^{x \ln (x)} (2^{x} \ln (2)) x + 2^{1 + x} x e^{x \ln (x)} \ln (x) + 2^{x} e^{x \ln (x)} x + 2^{x} e^{x \ln (x)} + 2^{x} (e^{x \ln (x)} \ln^{2} (x)) x + e^{x \ln (x)} (2^{x} \ln (2)) \ln (x) x}{x} + \ln (2) \cdot (2^{x} e^{x \ln (x)}) + \ln (2) \cdot (2^{x} (e^{x \ln (x)} \ln (x))) + x^{x} (2^{x} \ln^{2} (2))$

Paso 2.5.10.21

Para escribir $\ln (2) \cdot 2^{x} e^{x \ln (x)}$ como una fracción con un denominador común, multiplica por $\frac{x}{x}$ .

$f'' (x) = \frac{e^{x \ln (x)} (2^{x} \ln (2)) x + 2^{1 + x} x e^{x \ln (x)} \ln (x) + 2^{x} e^{x \ln (x)} x + 2^{x} e^{x \ln (x)} + 2^{x} (e^{x \ln (x)} \ln^{2} (x)) x + e^{x \ln (x)} (2^{x} \ln (2)) \ln (x) x}{x} + \ln (2) \cdot (2^{x} e^{x \ln (x)}) \cdot \frac{x}{x} + \ln (2) \cdot (2^{x} (e^{x \ln (x)} \ln (x))) + x^{x} (2^{x} \ln^{2} (2))$

Paso 2.5.10.22

Combina $\ln (2) \cdot 2^{x} e^{x \ln (x)}$ y $\frac{x}{x}$ .

$f'' (x) = \frac{e^{x \ln (x)} (2^{x} \ln (2)) x + 2^{1 + x} x e^{x \ln (x)} \ln (x) + 2^{x} e^{x \ln (x)} x + 2^{x} e^{x \ln (x)} + 2^{x} (e^{x \ln (x)} \ln^{2} (x)) x + e^{x \ln (x)} (2^{x} \ln (2)) \ln (x) x}{x} + \frac{\ln (2) \cdot (2^{x} e^{x \ln (x)} x)}{x} + \ln (2) \cdot (2^{x} (e^{x \ln (x)} \ln (x))) + x^{x} (2^{x} \ln^{2} (2))$

Paso 2.5.10.23

Combina los numeradores sobre el denominador común.

$f'' (x) = \frac{e^{x \ln (x)} (2^{x} \ln (2)) x + 2^{1 + x} x e^{x \ln (x)} \ln (x) + 2^{x} e^{x \ln (x)} x + 2^{x} e^{x \ln (x)} + 2^{x} (e^{x \ln (x)} \ln^{2} (x)) x + e^{x \ln (x)} (2^{x} \ln (2)) \ln (x) x + \ln (2) \cdot (2^{x} e^{x \ln (x)} x)}{x} + \ln (2) \cdot (2^{x} (e^{x \ln (x)} \ln (x))) + x^{x} (2^{x} \ln^{2} (2))$

Paso 2.5.10.24

Mueve $e^{x \ln (x)}$ .

$f'' (x) = \frac{(2^{x}) x e^{x \ln (x)} \ln (2) + 2^{x} x e^{x \ln (x)} \ln (2) + 2^{1 + x} x e^{x \ln (x)} \ln (x) + 2^{x} e^{x \ln (x)} x + 2^{x} e^{x \ln (x)} + 2^{x} (e^{x \ln (x)} \ln^{2} (x)) x + e^{x \ln (x)} (2^{x} \ln (2)) \ln (x) x}{x} + \ln (2) \cdot (2^{x} (e^{x \ln (x)} \ln (x))) + x^{x} (2^{x} \ln^{2} (2))$

Paso 2.5.10.25

Suma $(2^{x}) x e^{x \ln (x)} \ln (2)$ y $2^{x} x e^{x \ln (x)} \ln (2)$ .

$f'' (x) = \frac{2 \cdot (2^{x} x e^{x \ln (x)} \ln (2)) + 2^{1 + x} x e^{x \ln (x)} \ln (x) + 2^{x} e^{x \ln (x)} x + 2^{x} e^{x \ln (x)} + 2^{x} (e^{x \ln (x)} \ln^{2} (x)) x + e^{x \ln (x)} (2^{x} \ln (2)) \ln (x) x}{x} + \ln (2) \cdot (2^{x} (e^{x \ln (x)} \ln (x))) + x^{x} (2^{x} \ln^{2} (2))$

Paso 2.5.10.26

Multiplica $2$ por $2^{x}$ .

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Paso 2.5.10.26.1

Eleva $2$ a la potencia de $1$ .

Paso 2.5.10.26.2

Usa la regla de la potencia $a^{m} a^{n} = a^{m + n}$ para combinar exponentes.

$f'' (x) = \frac{2^{1 + x} x e^{x \ln (x)} \ln (2) + 2^{1 + x} x e^{x \ln (x)} \ln (x) + 2^{x} e^{x \ln (x)} x + 2^{x} e^{x \ln (x)} + 2^{x} (e^{x \ln (x)} \ln^{2} (x)) x + e^{x \ln (x)} (2^{x} \ln (2)) \ln (x) x}{x} + \ln (2) \cdot (2^{x} (e^{x \ln (x)} \ln (x))) + x^{x} (2^{x} \ln^{2} (2))$

Paso 2.5.10.27

Para escribir $\ln (2) \cdot 2^{x} (e^{x \ln (x)} \ln (x))$ como una fracción con un denominador común, multiplica por $\frac{x}{x}$ .

Paso 2.5.10.28

Combina $\ln (2) \cdot 2^{x} (e^{x \ln (x)} \ln (x))$ y $\frac{x}{x}$ .

$f'' (x) = \frac{2^{1 + x} x e^{x \ln (x)} \ln (2) + 2^{1 + x} x e^{x \ln (x)} \ln (x) + 2^{x} e^{x \ln (x)} x + 2^{x} e^{x \ln (x)} + 2^{x} (e^{x \ln (x)} \ln^{2} (x)) x + e^{x \ln (x)} (2^{x} \ln (2)) \ln (x) x}{x} + \frac{\ln (2) \cdot (2^{x} (e^{x \ln (x)} \ln (x)) x)}{x} + x^{x} (2^{x} \ln^{2} (2))$

Paso 2.5.10.29

Combina los numeradores sobre el denominador común.

$f'' (x) = \frac{2^{1 + x} x e^{x \ln (x)} \ln (2) + 2^{1 + x} x e^{x \ln (x)} \ln (x) + 2^{x} e^{x \ln (x)} x + 2^{x} e^{x \ln (x)} + 2^{x} (e^{x \ln (x)} \ln^{2} (x)) x + e^{x \ln (x)} (2^{x} \ln (2)) \ln (x) x + \ln (2) \cdot (2^{x} (e^{x \ln (x)} \ln (x)) x)}{x} + x^{x} (2^{x} \ln^{2} (2))$

Paso 2.5.10.30

Mueve $e^{x \ln (x)}$ .

$f'' (x) = \frac{2^{1 + x} x e^{x \ln (x)} \ln (2) + 2^{1 + x} x e^{x \ln (x)} \ln (x) + 2^{x} e^{x \ln (x)} x + 2^{x} e^{x \ln (x)} + 2^{x} (e^{x \ln (x)} \ln^{2} (x)) x + (2^{x}) x e^{x \ln (x)} \ln (2) \ln (x) + (2^{x}) x e^{x \ln (x)} \ln (2) \ln (x)}{x} + x^{x} (2^{x} \ln^{2} (2))$

Paso 2.5.10.31

Suma $(2^{x}) x e^{x \ln (x)} \ln (2) \ln (x)$ y $(2^{x}) x e^{x \ln (x)} \ln (2) \ln (x)$ .

$f'' (x) = \frac{2^{1 + x} x e^{x \ln (x)} \ln (2) + 2^{1 + x} x e^{x \ln (x)} \ln (x) + 2^{x} e^{x \ln (x)} x + 2^{x} e^{x \ln (x)} + 2^{x} (e^{x \ln (x)} \ln^{2} (x)) x + 2 \cdot (2^{x} x e^{x \ln (x)} \ln (2) \ln (x))}{x} + x^{x} (2^{x} \ln^{2} (2))$

Paso 2.5.10.32

Multiplica $2$ por $2^{x}$ .

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Paso 2.5.10.32.1

Eleva $2$ a la potencia de $1$ .

Paso 2.5.10.32.2

Usa la regla de la potencia $a^{m} a^{n} = a^{m + n}$ para combinar exponentes.

$f'' (x) = \frac{2^{1 + x} x e^{x \ln (x)} \ln (2) + 2^{1 + x} x e^{x \ln (x)} \ln (x) + 2^{x} e^{x \ln (x)} x + 2^{x} e^{x \ln (x)} + 2^{x} (e^{x \ln (x)} \ln^{2} (x)) x + 2^{1 + x} x e^{x \ln (x)} \ln (2) \ln (x)}{x} + x^{x} (2^{x} \ln^{2} (2))$

Paso 2.5.10.33

Para escribir $x^{x} (2^{x} \ln^{2} (2))$ como una fracción con un denominador común, multiplica por $\frac{x}{x}$ .

Paso 2.5.10.34

Combina los numeradores sobre el denominador común.

Paso 2.5.10.35

Eleva $x$ a la potencia de $1$ .

Paso 2.5.10.36

Usa la regla de la potencia $a^{m} a^{n} = a^{m + n}$ para combinar exponentes.

Paso 2.5.11

Reordena los términos.

$f'' (x) = \frac{2^{1 + x} e^{x \ln (x)} x \ln (2) + 2^{1 + x} e^{x \ln (x)} x \ln (x) + 2^{x} e^{x \ln (x)} x + 2^{x} e^{x \ln (x)} + 2^{x} e^{x \ln (x)} x \ln^{2} (x) + 2^{1 + x} e^{x \ln (x)} x \ln (2) \ln (x) + 2^{x} x^{1 + x} \ln^{2} (2)}{x}$

Paso 2.5.12

Reordena los factores en $\frac{2^{1 + x} e^{x \ln (x)} x \ln (2) + 2^{1 + x} e^{x \ln (x)} x \ln (x) + 2^{x} e^{x \ln (x)} x + 2^{x} e^{x \ln (x)} + 2^{x} e^{x \ln (x)} x \ln^{2} (x) + 2^{1 + x} e^{x \ln (x)} x \ln (2) \ln (x) + 2^{x} x^{1 + x} \ln^{2} (2)}{x}$ .

$f'' (x) = \frac{x \cdot (2^{1 + x} e^{x \ln (x)} \ln (2)) + x \cdot (2^{1 + x} e^{x \ln (x)} \ln (x)) + x \cdot (2^{x} e^{x \ln (x)}) + 2^{x} e^{x \ln (x)} + x \cdot (2^{x} e^{x \ln (x)} \ln^{2} (x)) + x \cdot (2^{1 + x} e^{x \ln (x)} \ln (2) \ln (x)) + 2^{x} x^{1 + x} \ln^{2} (2)}{x}$

Paso 3

Para obtener los valores mínimo y máximo locales de la función, establece la derivada igual a $0$ y resuelve.

$2^{x} e^{x \ln (x)} + 2^{x} e^{x \ln (x)} \ln (x) + 2^{x} x^{x} \ln (2) = 0$

Paso 4

Obtén la primera derivada.

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Paso 4.1

Obtén la primera derivada.

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Paso 4.1.1

Simplifica con la obtención del factor común.

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Paso 4.1.1.1

Factoriza $2$ de $2 x$ .

$\frac{d}{d x} [{(2 (x))}^{x}]$

Paso 4.1.1.2

Aplica la regla del producto a $2 (x)$ .

$\frac{d}{d x} [2^{x} x^{x}]$

Paso 4.1.2

Diferencia con la regla del producto, que establece que $\frac{d}{d x} [f (x) g (x)]$ es $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ donde $f (x) = 2^{x}$ y $g (x) = x^{x}$ .

$2^{x} \frac{d}{d x} [x^{x}] + x^{x} \frac{d}{d x} [2^{x}]$

Paso 4.1.3

Usa las propiedades de los logaritmos para simplificar la diferenciación.

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Paso 4.1.3.1

Reescribe $x^{x}$ como $e^{\ln (x^{x})}$ .

$2^{x} \frac{d}{d x} [e^{\ln (x^{x})}] + x^{x} \frac{d}{d x} [2^{x}]$

Paso 4.1.3.2

Expande $\ln (x^{x})$ ; para ello, mueve $x$ fuera del logaritmo.

$2^{x} \frac{d}{d x} [e^{x \ln (x)}] + x^{x} \frac{d}{d x} [2^{x}]$

Paso 4.1.4

Diferencia con la regla de la cadena, que establece que $\frac{d}{d x} [f (g (x))]$ es $f' (g (x)) g' (x)$ donde $f (x) = e^{x}$ y $g (x) = x \ln (x)$ .

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Paso 4.1.4.1

Para aplicar la regla de la cadena, establece $u$ como $x \ln (x)$ .

$2^{x} (\frac{d}{d u} [e^{u}] \frac{d}{d x} [x \ln (x)]) + x^{x} \frac{d}{d x} [2^{x}]$

Paso 4.1.4.2

Diferencia con la regla exponencial, que establece que $\frac{d}{d u} [a^{u}]$ es $a^{u} \ln (a)$ donde $a$ = $e$ .

$2^{x} (e^{u} \frac{d}{d x} [x \ln (x)]) + x^{x} \frac{d}{d x} [2^{x}]$

Paso 4.1.4.3

Reemplaza todos los casos de $u$ con $x \ln (x)$ .

$2^{x} (e^{x \ln (x)} \frac{d}{d x} [x \ln (x)]) + x^{x} \frac{d}{d x} [2^{x}]$

Paso 4.1.5

Diferencia con la regla del producto, que establece que $\frac{d}{d x} [f (x) g (x)]$ es $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ donde $f (x) = x$ y $g (x) = \ln (x)$ .

$2^{x} (e^{x \ln (x)} (x \frac{d}{d x} [\ln (x)] + \ln (x) \frac{d}{d x} [x])) + x^{x} \frac{d}{d x} [2^{x}]$

Paso 4.1.6

La derivada de $\ln (x)$ con respecto a $x$ es $\frac{1}{x}$ .

$2^{x} (e^{x \ln (x)} (x \frac{1}{x} + \ln (x) \frac{d}{d x} [x])) + x^{x} \frac{d}{d x} [2^{x}]$

Paso 4.1.7

Diferencia con la regla de la potencia.

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Paso 4.1.7.1

Combina $x$ y $\frac{1}{x}$ .

$2^{x} (e^{x \ln (x)} (\frac{x}{x} + \ln (x) \frac{d}{d x} [x])) + x^{x} \frac{d}{d x} [2^{x}]$

Paso 4.1.7.2

Cancela el factor común de $x$ .

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Paso 4.1.7.2.1

Cancela el factor común.

$2^{x} (e^{x \ln (x)} (\frac{x}{x} + \ln (x) \frac{d}{d x} [x])) + x^{x} \frac{d}{d x} [2^{x}]$

Paso 4.1.7.2.2

Reescribe la expresión.

$2^{x} (e^{x \ln (x)} (1 + \ln (x) \frac{d}{d x} [x])) + x^{x} \frac{d}{d x} [2^{x}]$

Paso 4.1.7.3

Diferencia con la regla de la potencia, que establece que $\frac{d}{d x} [x^{n}]$ es $n x^{n - 1}$ donde $n = 1$ .

$2^{x} (e^{x \ln (x)} (1 + \ln (x) \cdot 1)) + x^{x} \frac{d}{d x} [2^{x}]$

Paso 4.1.7.4

Multiplica $\ln (x)$ por $1$ .

$2^{x} (e^{x \ln (x)} (1 + \ln (x))) + x^{x} \frac{d}{d x} [2^{x}]$

Paso 4.1.8

Diferencia con la regla exponencial, que establece que $\frac{d}{d x} [a^{x}]$ es $a^{x} \ln (a)$ donde $a$ = $2$ .

$2^{x} (e^{x \ln (x)} (1 + \ln (x))) + x^{x} (2^{x} \ln (2))$

Paso 4.1.9

Simplifica.

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Paso 4.1.9.1

Aplica la propiedad distributiva.

$2^{x} (e^{x \ln (x)} \cdot 1 + e^{x \ln (x)} \ln (x)) + x^{x} (2^{x} \ln (2))$

Paso 4.1.9.2

Aplica la propiedad distributiva.

$2^{x} (e^{x \ln (x)} \cdot 1) + 2^{x} (e^{x \ln (x)} \ln (x)) + x^{x} (2^{x} \ln (2))$

Paso 4.1.9.3

Multiplica $e^{x \ln (x)}$ por $1$ .

$2^{x} e^{x \ln (x)} + 2^{x} (e^{x \ln (x)} \ln (x)) + x^{x} (2^{x} \ln (2))$

Paso 4.1.9.4

Reordena los términos.

$f' (x) = 2^{x} e^{x \ln (x)} + 2^{x} e^{x \ln (x)} \ln (x) + 2^{x} x^{x} \ln (2)$

Paso 4.2

La primera derivada de $f (x)$ con respecto a $x$ es $2^{x} e^{x \ln (x)} + 2^{x} e^{x \ln (x)} \ln (x) + 2^{x} x^{x} \ln (2)$ .

$2^{x} e^{x \ln (x)} + 2^{x} e^{x \ln (x)} \ln (x) + 2^{x} x^{x} \ln (2)$

Paso 5

Establece la primera derivada igual a $0$ , luego resuelve la ecuación $2^{x} e^{x \ln (x)} + 2^{x} e^{x \ln (x)} \ln (x) + 2^{x} x^{x} \ln (2) = 0$ .

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Paso 5.1

Establece la primera derivada igual a $0$ .

$2^{x} e^{x \ln (x)} + 2^{x} e^{x \ln (x)} \ln (x) + 2^{x} x^{x} \ln (2) = 0$

Paso 5.2

Grafica cada lado de la ecuación. La solución es el valor x del punto de intersección.

$x \approx 0.18393972$

Paso 6

Obtén los valores en el lugar donde la derivada es indefinida.

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Paso 6.1

Establece el argumento en $\ln (x)$ menor o igual que $0$ para obtener el lugar donde no está definida la expresión.

$x \leq 0$

Paso 6.2

La ecuación es indefinida cuando el denominador es igual a $0$ , el argumento de una raíz cuadrada es menor que $0$ o el argumento de un logaritmo es menor o igual que $0$ .

$x \leq 0$

$(- \infty, 0]$

$x \leq 0$

$(- \infty, 0]$

Paso 7

Puntos críticos para evaluar.

$x = 0.18393972$

Paso 8

Evalúa la segunda derivada en $x = 0.18393972$ . Si la segunda derivada es positiva, entonces este es un mínimo local. Si es negativa, entonces este es un máximo local.

$\frac{0.18393972 \cdot (2^{1 + 0.18393972} e^{(0.18393972) \ln (0.18393972)} \ln (2)) + 0.18393972 \cdot (2^{1 + 0.18393972} e^{(0.18393972) \ln (0.18393972)} \ln (0.18393972)) + (0.18393972) \cdot (2^{0.18393972} e^{(0.18393972) \ln (0.18393972)}) + 2^{0.18393972} e^{(0.18393972) \ln (0.18393972)} + (0.18393972) \cdot (2^{0.18393972} e^{(0.18393972) \ln (0.18393972)} \ln^{2} (0.18393972)) + 0.18393972 \cdot (2^{1 + 0.18393972} e^{(0.18393972) \ln (0.18393972)} \ln (2) \ln (0.18393972)) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9

Evalúa la segunda derivada.

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Paso 9.1

Elimina los paréntesis.

Paso 9.2

Simplifica el numerador.

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Paso 9.2.1

Suma $1$ y $0.18393972$ .

$\frac{0.18393972 \cdot 2^{1.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (0.18393972) + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} \ln^{2} (0.18393972) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.2

Eleva $2$ a la potencia de $1.18393972$ .

$\frac{0.18393972 \cdot 2.27196359 e^{0.18393972 \ln (0.18393972)} \ln (2) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (0.18393972) + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} \ln^{2} (0.18393972) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.3

Multiplica $0.18393972$ por $2.27196359$ .

$\frac{0.41790434 e^{0.18393972 \ln (0.18393972)} \ln (2) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (0.18393972) + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} \ln^{2} (0.18393972) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.4

Simplifica $0.18393972 \ln (0.18393972)$ al mover $0.18393972$ dentro del algoritmo.

$\frac{0.41790434 e^{\ln ({0.18393972}^{0.18393972})} \ln (2) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (0.18393972) + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} \ln^{2} (0.18393972) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.5

Potencia y logaritmo son funciones inversas.

$\frac{0.41790434 \cdot {0.18393972}^{0.18393972} \ln (2) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (0.18393972) + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} \ln^{2} (0.18393972) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.6

Eleva $0.18393972$ a la potencia de $0.18393972$ .

$\frac{0.41790434 \cdot 0.73239373 \ln (2) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (0.18393972) + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} \ln^{2} (0.18393972) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.7

Multiplica $0.41790434$ por $0.73239373$ .

$\frac{0.30607052 \ln (2) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (0.18393972) + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} \ln^{2} (0.18393972) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.8

Simplifica $0.30607052 \ln (2)$ al mover $0.30607052$ dentro del algoritmo.

$\frac{\ln (2^{0.30607052}) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (0.18393972) + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} \ln^{2} (0.18393972) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.9

Eleva $2$ a la potencia de $0.30607052$ .

$\frac{\ln (1.23633569) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (0.18393972) + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} \ln^{2} (0.18393972) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.10

Suma $1$ y $0.18393972$ .

$\frac{\ln (1.23633569) + 0.18393972 \cdot 2^{1.18393972} e^{0.18393972 \ln (0.18393972)} \ln (0.18393972) + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} \ln^{2} (0.18393972) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.11

Eleva $2$ a la potencia de $1.18393972$ .

$\frac{\ln (1.23633569) + 0.18393972 \cdot 2.27196359 e^{0.18393972 \ln (0.18393972)} \ln (0.18393972) + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} \ln^{2} (0.18393972) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.12

Multiplica $0.18393972$ por $2.27196359$ .

$\frac{\ln (1.23633569) + 0.41790434 e^{0.18393972 \ln (0.18393972)} \ln (0.18393972) + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} \ln^{2} (0.18393972) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.13

Simplifica $0.18393972 \ln (0.18393972)$ al mover $0.18393972$ dentro del algoritmo.

$\frac{\ln (1.23633569) + 0.41790434 e^{\ln ({0.18393972}^{0.18393972})} \ln (0.18393972) + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} \ln^{2} (0.18393972) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.14

Potencia y logaritmo son funciones inversas.

$\frac{\ln (1.23633569) + 0.41790434 \cdot {0.18393972}^{0.18393972} \ln (0.18393972) + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} \ln^{2} (0.18393972) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.15

Eleva $0.18393972$ a la potencia de $0.18393972$ .

$\frac{\ln (1.23633569) + 0.41790434 \cdot 0.73239373 \ln (0.18393972) + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} \ln^{2} (0.18393972) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.16

Multiplica $0.41790434$ por $0.73239373$ .

$\frac{\ln (1.23633569) + 0.30607052 \ln (0.18393972) + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} \ln^{2} (0.18393972) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.17

Simplifica $0.30607052 \ln (0.18393972)$ al mover $0.30607052$ dentro del algoritmo.

$\frac{\ln (1.23633569) + \ln ({0.18393972}^{0.30607052}) + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} \ln^{2} (0.18393972) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.18

Eleva $0.18393972$ a la potencia de $0.30607052$ .

$\frac{\ln (1.23633569) + \ln (0.59557827) + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} \ln^{2} (0.18393972) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.19

Eleva $2$ a la potencia de $0.18393972$ .

$\frac{\ln (1.23633569) + \ln (0.59557827) + 0.18393972 \cdot 1.13598179 e^{0.18393972 \ln (0.18393972)} + 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} \ln^{2} (0.18393972) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.20

Multiplica $0.18393972$ por $1.13598179$ .

$\frac{\ln (1.23633569) + \ln (0.59557827) + 0.20895217 e^{0.18393972 \ln (0.18393972)} + 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} \ln^{2} (0.18393972) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.21

Simplifica $0.18393972 \ln (0.18393972)$ al mover $0.18393972$ dentro del algoritmo.

$\frac{\ln (1.23633569) + \ln (0.59557827) + 0.20895217 e^{\ln ({0.18393972}^{0.18393972})} + 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} \ln^{2} (0.18393972) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.22

Potencia y logaritmo son funciones inversas.

$\frac{\ln (1.23633569) + \ln (0.59557827) + 0.20895217 \cdot {0.18393972}^{0.18393972} + 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} \ln^{2} (0.18393972) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.23

Eleva $0.18393972$ a la potencia de $0.18393972$ .

$\frac{\ln (1.23633569) + \ln (0.59557827) + 0.20895217 \cdot 0.73239373 + 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} \ln^{2} (0.18393972) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.24

Multiplica $0.20895217$ por $0.73239373$ .

$\frac{\ln (1.23633569) + \ln (0.59557827) + 0.15303526 + 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} \ln^{2} (0.18393972) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.25

Eleva $2$ a la potencia de $0.18393972$ .

$\frac{\ln (1.23633569) + \ln (0.59557827) + 0.15303526 + 1.13598179 e^{0.18393972 \ln (0.18393972)} + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} \ln^{2} (0.18393972) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.26

Simplifica $0.18393972 \ln (0.18393972)$ al mover $0.18393972$ dentro del algoritmo.

$\frac{\ln (1.23633569) + \ln (0.59557827) + 0.15303526 + 1.13598179 e^{\ln ({0.18393972}^{0.18393972})} + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} \ln^{2} (0.18393972) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.27

Potencia y logaritmo son funciones inversas.

$\frac{\ln (1.23633569) + \ln (0.59557827) + 0.15303526 + 1.13598179 \cdot {0.18393972}^{0.18393972} + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} \ln^{2} (0.18393972) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.28

Eleva $0.18393972$ a la potencia de $0.18393972$ .

$\frac{\ln (1.23633569) + \ln (0.59557827) + 0.15303526 + 1.13598179 \cdot 0.73239373 + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} \ln^{2} (0.18393972) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.29

Multiplica $1.13598179$ por $0.73239373$ .

$\frac{\ln (1.23633569) + \ln (0.59557827) + 0.15303526 + 0.83198595 + 0.18393972 \cdot 2^{0.18393972} e^{0.18393972 \ln (0.18393972)} \ln^{2} (0.18393972) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.30

Eleva $2$ a la potencia de $0.18393972$ .

$\frac{\ln (1.23633569) + \ln (0.59557827) + 0.15303526 + 0.83198595 + 0.18393972 \cdot 1.13598179 e^{0.18393972 \ln (0.18393972)} \ln^{2} (0.18393972) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.31

Multiplica $0.18393972$ por $1.13598179$ .

$\frac{\ln (1.23633569) + \ln (0.59557827) + 0.15303526 + 0.83198595 + 0.20895217 e^{0.18393972 \ln (0.18393972)} \ln^{2} (0.18393972) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.32

Simplifica $0.18393972 \ln (0.18393972)$ al mover $0.18393972$ dentro del algoritmo.

$\frac{\ln (1.23633569) + \ln (0.59557827) + 0.15303526 + 0.83198595 + 0.20895217 e^{\ln ({0.18393972}^{0.18393972})} \ln^{2} (0.18393972) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.33

Potencia y logaritmo son funciones inversas.

$\frac{\ln (1.23633569) + \ln (0.59557827) + 0.15303526 + 0.83198595 + 0.20895217 \cdot {0.18393972}^{0.18393972} \ln^{2} (0.18393972) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.34

Eleva $0.18393972$ a la potencia de $0.18393972$ .

$\frac{\ln (1.23633569) + \ln (0.59557827) + 0.15303526 + 0.83198595 + 0.20895217 \cdot 0.73239373 \ln^{2} (0.18393972) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.35

Multiplica $0.20895217$ por $0.73239373$ .

$\frac{\ln (1.23633569) + \ln (0.59557827) + 0.15303526 + 0.83198595 + 0.15303526 \ln^{2} (0.18393972) + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.36

Multiplica $0.15303526$ por $\ln^{2} (0.18393972)$ .

$\frac{\ln (1.23633569) + \ln (0.59557827) + 0.15303526 + 0.83198595 + 0.43871344 + 0.18393972 \cdot 2^{1 + 0.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.37

Suma $1$ y $0.18393972$ .

$\frac{\ln (1.23633569) + \ln (0.59557827) + 0.15303526 + 0.83198595 + 0.43871344 + 0.18393972 \cdot 2^{1.18393972} e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.38

Eleva $2$ a la potencia de $1.18393972$ .

$\frac{\ln (1.23633569) + \ln (0.59557827) + 0.15303526 + 0.83198595 + 0.43871344 + 0.18393972 \cdot 2.27196359 e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.39

Multiplica $0.18393972$ por $2.27196359$ .

$\frac{\ln (1.23633569) + \ln (0.59557827) + 0.15303526 + 0.83198595 + 0.43871344 + 0.41790434 e^{0.18393972 \ln (0.18393972)} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.40

Simplifica $0.18393972 \ln (0.18393972)$ al mover $0.18393972$ dentro del algoritmo.

$\frac{\ln (1.23633569) + \ln (0.59557827) + 0.15303526 + 0.83198595 + 0.43871344 + 0.41790434 e^{\ln ({0.18393972}^{0.18393972})} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.41

Potencia y logaritmo son funciones inversas.

$\frac{\ln (1.23633569) + \ln (0.59557827) + 0.15303526 + 0.83198595 + 0.43871344 + 0.41790434 \cdot {0.18393972}^{0.18393972} \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.42

Eleva $0.18393972$ a la potencia de $0.18393972$ .

$\frac{\ln (1.23633569) + \ln (0.59557827) + 0.15303526 + 0.83198595 + 0.43871344 + 0.41790434 \cdot 0.73239373 \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.43

Multiplica $0.41790434$ por $0.73239373$ .

$\frac{\ln (1.23633569) + \ln (0.59557827) + 0.15303526 + 0.83198595 + 0.43871344 + 0.30607052 \ln (2) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.44

Simplifica $0.30607052 \ln (2)$ al mover $0.30607052$ dentro del algoritmo.

$\frac{\ln (1.23633569) + \ln (0.59557827) + 0.15303526 + 0.83198595 + 0.43871344 + \ln (2^{0.30607052}) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.45

Eleva $2$ a la potencia de $0.30607052$ .

$\frac{\ln (1.23633569) + \ln (0.59557827) + 0.15303526 + 0.83198595 + 0.43871344 + \ln (1.23633569) \ln (0.18393972) + 2^{0.18393972} \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.46

Eleva $2$ a la potencia de $0.18393972$ .

$\frac{\ln (1.23633569) + \ln (0.59557827) + 0.15303526 + 0.83198595 + 0.43871344 + \ln (1.23633569) \ln (0.18393972) + 1.13598179 \cdot {0.18393972}^{1 + 0.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.47

Suma $1$ y $0.18393972$ .

$\frac{\ln (1.23633569) + \ln (0.59557827) + 0.15303526 + 0.83198595 + 0.43871344 + \ln (1.23633569) \ln (0.18393972) + 1.13598179 \cdot {0.18393972}^{1.18393972} \ln^{2} (2)}{0.18393972}$

Paso 9.2.48

Eleva $0.18393972$ a la potencia de $1.18393972$ .

$\frac{\ln (1.23633569) + \ln (0.59557827) + 0.15303526 + 0.83198595 + 0.43871344 + \ln (1.23633569) \ln (0.18393972) + 1.13598179 \cdot 0.13471629 \ln^{2} (2)}{0.18393972}$

Paso 9.2.49

Multiplica $1.13598179$ por $0.13471629$ .

$\frac{\ln (1.23633569) + \ln (0.59557827) + 0.15303526 + 0.83198595 + 0.43871344 + \ln (1.23633569) \ln (0.18393972) + 0.15303526 \ln^{2} (2)}{0.18393972}$

Paso 9.2.50

Multiplica $0.15303526$ por $\ln^{2} (2)$ .

$\frac{\ln (1.23633569) + \ln (0.59557827) + 0.15303526 + 0.83198595 + 0.43871344 + \ln (1.23633569) \ln (0.18393972) + 0.07352625}{0.18393972}$

Paso 9.2.51

Usa las propiedades de los logaritmos del producto, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\frac{\ln (1.23633569 \cdot 0.59557827) + 0.15303526 + 0.83198595 + 0.43871344 + \ln (1.23633569) \ln (0.18393972) + 0.07352625}{0.18393972}$

Paso 9.2.52

Suma $\ln (1.23633569 \cdot 0.59557827)$ y $0.15303526$ .

$\frac{- 0.15303526 + 0.83198595 + 0.43871344 + \ln (1.23633569) \ln (0.18393972) + 0.07352625}{0.18393972}$

Paso 9.2.53

Suma $- 0.15303526$ y $0.83198595$ .

$\frac{0.67895069 + 0.43871344 + \ln (1.23633569) \ln (0.18393972) + 0.07352625}{0.18393972}$

Paso 9.2.54

Suma $0.67895069$ y $0.43871344$ .

$\frac{1.11766413 + \ln (1.23633569) \ln (0.18393972) + 0.07352625}{0.18393972}$

Paso 9.2.55

Suma $1.11766413$ y $\ln (1.23633569) \ln (0.18393972)$ .

$\frac{0.7584597 + 0.07352625}{0.18393972}$

Paso 9.2.56

Suma $0.7584597$ y $0.07352625$ .

$\frac{0.83198595}{0.18393972}$

Paso 9.3

Divide $0.83198595$ por $0.18393972$ .

$4.52314459$

Paso 10

$x = 0.18393972$ es un mínimo local porque el valor de la segunda derivada es positivo. Esto se conoce como prueba de la segunda derivada.

$x = 0.18393972$ es un mínimo local

Paso 11

Obtén el valor de y cuando $x = 0.18393972$ .

Toca para ver más pasos...

Paso 11.1

Reemplaza la variable $x$ con $0.18393972$ en la expresión.

$f (0.18393972) = {(2 (0.18393972))}^{0.18393972}$

Paso 11.2

Simplifica el resultado.

Toca para ver más pasos...

Paso 11.2.1

Multiplica $2$ por $0.18393972$ .

$f (0.18393972) = {0.36787944}^{0.18393972}$

Paso 11.2.2

Eleva $0.36787944$ a la potencia de $0.18393972$ .

$f (0.18393972) = 0.83198595$

Paso 11.2.3

La respuesta final es $0.83198595$ .

$y = 0.83198595$

Paso 12

Estos son los extremos locales de $f (x) = {(2 x)}^{x}$ .

$(0.18393972, 0.83198595)$ es un mínimo local

Paso 13