Calculus Examples

$y = x^{1 - x}$

Step 1

Use the properties of logarithms to simplify the differentiation.

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Step 1.1

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

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

Step 1.2

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

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

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 $g (x) = (1 - x) \ln (x)$ .

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Step 2.1

To apply the Chain Rule, set $u$ as $(1 - x) \ln (x)$ .

$\frac{d}{d u} [e^{u}] \frac{d}{d x} [(1 - x) \ln (x)]$

Step 2.2

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

$e^{u} \frac{d}{d x} [(1 - x) \ln (x)]$

Step 2.3

Replace all occurrences of $u$ with $(1 - x) \ln (x)$ .

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

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 $f (x) = 1 - x$ and $g (x) = \ln (x)$ .

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

Step 4

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

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

Step 5

Differentiate.

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Step 5.1

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

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

Step 5.2

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

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

Step 5.3

Add $0$ and $\frac{d}{d x} [- x]$ .

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

Step 5.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

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

Step 5.5

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

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

Step 5.6

Simplify the expression.

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Step 5.6.1

Multiply $- 1$ by $1$ .

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

Step 5.6.2

Move $- 1$ to the left of $\ln (x)$ .

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

Step 5.6.3

Rewrite $- 1 \ln (x)$ as $- \ln (x)$ .

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

Step 6

Simplify.

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Step 6.1

Apply the distributive property.

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

Step 6.2

Apply the distributive property.

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

Step 6.3

Combine terms.

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Step 6.3.1

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

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

Step 6.3.2

Multiply $\frac{1}{x}$ by $1$ .

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

Step 6.3.3

Combine $\frac{1}{x}$ and $x$ .

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

Step 6.3.4

Cancel the common factor of $x$ .

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Step 6.3.4.1

Cancel the common factor.

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

Step 6.3.4.2

Rewrite the expression.

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

Step 6.3.5

Multiply $- 1$ by $1$ .

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

Step 6.4

Apply the distributive property.

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

Step 6.5

Simplify.

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Step 6.5.1

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

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

Step 6.5.2

Move $- 1$ to the left of $e^{\ln (x) - x \ln (x)}$ .

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

Step 6.5.3

Rewrite using the commutative property of multiplication.

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

Step 6.6

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

$\frac{e^{\ln (x) - x \ln (x)}}{x} - e^{\ln (x) - x \ln (x)} - e^{\ln (x) - x \ln (x)} \ln (x)$

Step 6.7

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

$\frac{e^{\ln (x) - x \ln (x)}}{x} - e^{\ln (x) - x \ln (x)} \cdot \frac{x}{x} - e^{\ln (x) - x \ln (x)} \ln (x)$

Step 6.8

Combine $- e^{\ln (x) - x \ln (x)}$ and $\frac{x}{x}$ .

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

Step 6.9

Combine the numerators over the common denominator.

$\frac{e^{\ln (x) - x \ln (x)} - e^{\ln (x) - x \ln (x)} x}{x} - e^{\ln (x) - x \ln (x)} \ln (x)$

Step 6.10

Factor $e^{\ln (x) - x \ln (x)}$ out of $e^{\ln (x) - x \ln (x)} - e^{\ln (x) - x \ln (x)} x$ .

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Step 6.10.1

Multiply by $1$ .

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

Step 6.10.2

Factor $e^{\ln (x) - x \ln (x)}$ out of $- e^{\ln (x) - x \ln (x)} x$ .

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

Step 6.10.3

Factor $e^{\ln (x) - x \ln (x)}$ out of $e^{\ln (x) - x \ln (x)} \cdot 1 + e^{\ln (x) - x \ln (x)} (- x)$ .

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

Step 6.11

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

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

Step 6.12

Combine $- e^{\ln (x) - x \ln (x)} \ln (x)$ and $\frac{x}{x}$ .

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

Step 6.13

Combine the numerators over the common denominator.

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

Step 6.14

Factor $e^{\ln (x) - x \ln (x)}$ out of $e^{\ln (x) - x \ln (x)} (1 - x) - e^{\ln (x) - x \ln (x)} \ln (x) x$ .

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Step 6.14.1

Factor $e^{\ln (x) - x \ln (x)}$ out of $- e^{\ln (x) - x \ln (x)} \ln (x) x$ .

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

Step 6.14.2

Factor $e^{\ln (x) - x \ln (x)}$ out of $e^{\ln (x) - x \ln (x)} (1 - x) + e^{\ln (x) - x \ln (x)} (- \ln (x) x)$ .

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

Step 6.15

Reorder factors in $\frac{e^{\ln (x) - x \ln (x)} (1 - x - \ln (x) x)}{x}$ .

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