Calculus Examples

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

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 $x$ outside the logarithm.

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

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

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Step 2.1

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

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

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 $a$ = $e$ .

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

Step 2.3

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

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

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

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

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 $g (x) = x + 1$ .

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Step 4.1

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

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

Step 4.2

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

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

Step 4.3

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

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

Step 5

Differentiate.

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Step 5.1

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

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

Step 5.2

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

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

Step 5.3

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

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

Step 5.4

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

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

Step 5.5

Simplify the expression.

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Step 5.5.1

Add $1$ and $0$ .

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

Step 5.5.2

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

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

Step 5.6

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

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

Step 5.7

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

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

Step 6

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

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

Step 7

Combine the numerators over the common denominator.

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

Step 8

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

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

Step 9

Simplify.

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Step 9.1

Simplify the numerator.

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Step 9.1.1

Simplify each term.

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Step 9.1.1.1

Apply the distributive property.

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

Step 9.1.1.2

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

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

Step 9.1.2

Apply the distributive property.

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

Step 9.1.3

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

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

Step 9.2

Reorder terms.

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