Calculus Examples

$\ln (\ln (1 - x))$

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

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Step 1.1

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

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

Step 1.2

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

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

Step 1.3

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

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

Step 2

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

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Step 2.1

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

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

Step 2.2

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

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

Step 2.3

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

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

Step 3

Differentiate.

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Step 3.1

Multiply $\frac{1}{1 - x}$ by $\frac{1}{\ln (1 - x)}$ .

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

Step 3.2

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

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

Step 3.3

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

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

Step 3.4

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

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

Step 3.5

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

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

Step 3.6

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

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

Step 3.7

Combine fractions.

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Step 3.7.1

Multiply $- 1$ by $1$ .

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

Step 3.7.2

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

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

Step 3.7.3

Simplify the expression.

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Step 3.7.3.1

Move the negative in front of the fraction.

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

Step 3.7.3.2

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

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