Calculus Examples

${(x^{2} + 1)}^{2^{x}}$

Step 1

Use the properties of logarithms to simplify the differentiation.

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Step 1.1

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

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

Step 1.2

Expand $\ln ({(x^{2} + 1)}^{2^{x}})$ by moving $2^{x}$ outside the logarithm.

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

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = 2^{x} \ln (x^{2} + 1)$ .

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Step 2.1

To apply the Chain Rule, set $u_{1}$ as $2^{x} \ln (x^{2} + 1)$ .

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

Step 2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

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

Step 2.3

Replace all occurrences of $u_{1}$ with $2^{x} \ln (x^{2} + 1)$ .

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

Step 3

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = 2^{x}$ and $g (x) = \ln (x^{2} + 1)$ .

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

Step 4

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = x^{2} + 1$ .

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Step 4.1

To apply the Chain Rule, set $u_{2}$ as $x^{2} + 1$ .

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

Step 4.2

The derivative of $\ln (u_{2})$ with respect to $u_{2}$ is $\frac{1}{u_{2}}$ .

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

Step 4.3

Replace all occurrences of $u_{2}$ with $x^{2} + 1$ .

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

Step 5

Differentiate.

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Step 5.1

Combine $\frac{1}{x^{2} + 1}$ and $2^{x}$ .

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

Step 5.2

By the Sum Rule, the derivative of $x^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]$ .

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

Step 5.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

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

Step 5.4

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$e^{2^{x} \ln (x^{2} + 1)} (\frac{2^{x}}{x^{2} + 1} (2 x + 0) + \ln (x^{2} + 1) \frac{d}{d x} [2^{x}])$

Step 5.5

Combine fractions.

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Step 5.5.1

Add $2 x$ and $0$ .

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

Step 5.5.2

Combine $2$ and $\frac{2^{x}}{x^{2} + 1}$ .

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

Step 6

Multiply $2$ by $2^{x}$ .

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Step 6.1

Raise $2$ to the power of $1$ .

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

Step 6.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 7

Combine $\frac{2^{1 + x}}{x^{2} + 1}$ and $x$ .

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

Step 8

Differentiate using the Exponential Rule which states that $\frac{d}{d x} [a^{x}]$ is $a^{x} \ln (a)$ where $a$ = $2$ .

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

Step 9

To write $\ln (x^{2} + 1) (2^{x} \ln (2))$ as a fraction with a common denominator, multiply by $\frac{x^{2} + 1}{x^{2} + 1}$ .

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

Step 10

Combine the numerators over the common denominator.

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

Step 11

Combine $e^{2^{x} \ln (x^{2} + 1)}$ and $\frac{2^{1 + x} x + \ln (x^{2} + 1) (2^{x} \ln (2)) (x^{2} + 1)}{x^{2} + 1}$ .

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

Step 12

Simplify.

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Step 12.1

Simplify the numerator.

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Step 12.1.1

Simplify each term.

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Step 12.1.1.1

Rewrite using the commutative property of multiplication.

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

Step 12.1.1.2

Apply the distributive property.

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

Step 12.1.1.3

Multiply $\ln (2)$ by $1$ .

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

Step 12.1.2

Apply the distributive property.

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

Step 12.1.3

Simplify.

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Step 12.1.3.1

Rewrite using the commutative property of multiplication.

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

Step 12.1.3.2

Rewrite using the commutative property of multiplication.

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

Step 12.1.4

Reorder factors in $e^{2^{x} \ln (x^{2} + 1)} (2^{1 + x} x) + \ln (2) e^{2^{x} \ln (x^{2} + 1)} (\ln (x^{2} + 1) \cdot 2^{x} x^{2}) + \ln (2) e^{2^{x} \ln (x^{2} + 1)} (\ln (x^{2} + 1) \cdot 2^{x})$ .

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

Step 12.2

Reorder terms.

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