Calculus Examples

$y = {(1 + x)}^{\frac{1}{x}}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (y) = \frac{d}{d x} ({(1 + x)}^{\frac{1}{x}})$

Step 2

The derivative of $y$ with respect to $x$ is $y'$ .

$y'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

Use the properties of logarithms to simplify the differentiation.

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Step 3.1.1

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

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

Step 3.1.2

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

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

Step 3.2

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

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

Step 3.3

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 $g (x) = \frac{\ln (1 + x)}{x}$ .

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Step 3.3.1

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

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

Step 3.3.2

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

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

Step 3.3.3

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

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

Step 3.4

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

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

Step 3.5

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 $g (x) = 1 + x$ .

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Step 3.5.1

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

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

Step 3.5.2

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

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

Step 3.5.3

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

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

Step 3.6

Differentiate.

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Step 3.6.1

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

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

Step 3.6.2

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

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

Step 3.6.3

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

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

Step 3.6.4

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

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

Step 3.6.5

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

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

Step 3.6.6

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

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

Step 3.7

Multiply $\frac{\frac{x}{1 + x} - \ln (1 + x) \frac{d}{d x} [x]}{x^{2}}$ by $\frac{1 + x}{1 + x}$ .

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

Step 3.8

Simplify terms.

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Step 3.8.1

Combine.

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

Step 3.8.2

Apply the distributive property.

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

Step 3.8.3

Cancel the common factor of $1 + x$ .

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Step 3.8.3.1

Cancel the common factor.

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

Step 3.8.3.2

Rewrite the expression.

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

Step 3.9

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

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

Step 3.10

Combine fractions.

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Step 3.10.1

Multiply $- 1$ by $1$ .

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

Step 3.10.2

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

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

Step 3.11

Simplify.

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Step 3.11.1

Apply the distributive property.

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

Step 3.11.2

Simplify the numerator.

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Step 3.11.2.1

Simplify each term.

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Step 3.11.2.1.1

Apply the distributive property.

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

Step 3.11.2.1.2

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

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

Step 3.11.2.1.3

Rewrite using the commutative property of multiplication.

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

Step 3.11.2.2

Apply the distributive property.

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

Step 3.11.2.3

Simplify.

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Step 3.11.2.3.1

Rewrite using the commutative property of multiplication.

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

Step 3.11.2.3.2

Rewrite using the commutative property of multiplication.

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

Step 3.11.2.4

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

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

Step 3.11.3

Combine terms.

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Step 3.11.3.1

Multiply $x^{2}$ by $1$ .

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

Step 3.11.3.2

Multiply $x$ by $x^{2}$ by adding the exponents.

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Step 3.11.3.2.1

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

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Step 3.11.3.2.1.1

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

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

Step 3.11.3.2.1.2

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

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

Step 3.11.3.2.2

Add $1$ and $2$ .

$\frac{x e^{\frac{\ln (1 + x)}{x}} - e^{\frac{\ln (1 + x)}{x}} \ln (1 + x) - x e^{\frac{\ln (1 + x)}{x}} \ln (1 + x)}{x^{2} + x^{3}}$

Step 3.11.4

Reorder terms.

$\frac{e^{\frac{\ln (1 + x)}{x}} x - e^{\frac{\ln (1 + x)}{x}} \ln (1 + x) - e^{\frac{\ln (1 + x)}{x}} x \ln (1 + x)}{x^{3} + x^{2}}$

Step 4

Reform the equation by setting the left side equal to the right side.

$y' = \frac{e^{\frac{\ln (1 + x)}{x}} x - e^{\frac{\ln (1 + x)}{x}} \ln (1 + x) - e^{\frac{\ln (1 + x)}{x}} x \ln (1 + x)}{x^{3} + x^{2}}$

Step 5

Replace $y'$ with $\frac{d y}{d x}$ .

$\frac{d y}{d x} = \frac{e^{\frac{\ln (1 + x)}{x}} x - e^{\frac{\ln (1 + x)}{x}} \ln (1 + x) - e^{\frac{\ln (1 + x)}{x}} x \ln (1 + x)}{x^{3} + x^{2}}$