Calculus Examples

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

Step 1

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

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Step 1.1

To apply the Chain Rule, set $u$ as $\frac{x}{x + 1}$ .

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

Step 1.2

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

$\frac{1}{u} \frac{d}{d x} [\frac{x}{x + 1}]$

Step 1.3

Replace all occurrences of $u$ with $\frac{x}{x + 1}$ .

$\frac{1}{\frac{x}{x + 1}} \frac{d}{d x} [\frac{x}{x + 1}]$

Step 2

Multiply by the reciprocal of the fraction to divide by $\frac{x}{x + 1}$ .

$1 \frac{x + 1}{x} \frac{d}{d x} [\frac{x}{x + 1}]$

Step 3

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

$\frac{x + 1}{x} \frac{d}{d x} [\frac{x}{x + 1}]$

Step 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 $f (x) = x$ and $g (x) = x + 1$ .

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

Step 5

Differentiate.

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Step 5.1

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

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

Step 5.2

Multiply $x + 1$ by $1$ .

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

Step 5.3

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

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

Step 5.4

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

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

Step 5.5

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

$\frac{x + 1}{x} \cdot \frac{x + 1 - x (1 + 0)}{{(x + 1)}^{2}}$

Step 5.6

Simplify terms.

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Step 5.6.1

Add $1$ and $0$ .

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

Step 5.6.2

Multiply $- 1$ by $1$ .

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

Step 5.6.3

Subtract $x$ from $x$ .

$\frac{x + 1}{x} \cdot \frac{0 + 1}{{(x + 1)}^{2}}$

Step 5.6.4

Add $0$ and $1$ .

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

Step 5.6.5

Multiply $\frac{x + 1}{x}$ by $\frac{1}{{(x + 1)}^{2}}$ .

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

Step 5.6.6

Cancel the common factor of $x + 1$ and ${(x + 1)}^{2}$ .

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Step 5.6.6.1

Multiply by $1$ .

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

Step 5.6.6.2

Cancel the common factors.

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Step 5.6.6.2.1

Factor $x + 1$ out of $x {(x + 1)}^{2}$ .

$\frac{(x + 1) \cdot 1}{(x + 1) (x (x + 1))}$

Step 5.6.6.2.2

Cancel the common factor.

$\frac{(x + 1) \cdot 1}{(x + 1) (x (x + 1))}$

Step 5.6.6.2.3

Rewrite the expression.

$\frac{1}{x (x + 1)}$