Calculus Examples

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

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

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Step 1.1

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

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

Step 1.2

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

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

Step 1.3

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

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

Step 2

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

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

Step 3

Simplify the expression.

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Step 3.1

Multiply $x + 1$ by $1$ .

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

Step 3.2

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

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

Step 4

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

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Step 4.1

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

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

Step 4.2

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

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

Step 4.3

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

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

Step 5

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

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

Step 6

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

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

Step 7

Subtract $2$ from $1$ .

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

Step 8

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

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

Step 9

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

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

Step 10

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

$- {(x + 1)}^{- 1} (1 + 0)$

Step 11

Simplify the expression.

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Step 11.1

Add $1$ and $0$ .

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

Step 11.2

Multiply $- 1$ by $1$ .

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

Step 12

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$- \frac{1}{x + 1}$