Calculus Examples

$\ln (| 1 - x |)$

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 $g (x) = | 1 - x |$ .

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Step 1.1

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

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

Step 1.2

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

$\frac{1}{u_{1}} \frac{d}{d x} [| 1 - x |]$

Step 1.3

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

$\frac{1}{| 1 - x |} \frac{d}{d x} [| 1 - x |]$

Step 2

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

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Step 2.1

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

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

Step 2.2

The derivative of $| u_{2} |$ with respect to $u_{2}$ is $\frac{u_{2}}{| u_{2} |}$ .

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

Step 2.3

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

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

Step 3

Multiply $\frac{1 - x}{| 1 - x |}$ by $\frac{1}{| 1 - x |}$ .

$\frac{1 - x}{| 1 - x | | 1 - x |} \frac{d}{d x} [1 - x]$

Step 4

To multiply absolute values, multiply the terms inside each absolute value.

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

Step 5

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

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

Step 6

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

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

Step 7

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

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

Step 8

Add $1$ and $1$ .

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

Step 9

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

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

Step 10

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

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

Step 11

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

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

Step 12

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

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

Step 13

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

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

Step 14

Combine fractions.

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Step 14.1

Multiply $- 1$ by $1$ .

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

Step 14.2

Combine $\frac{1 - x}{| {(1 - x)}^{2} |}$ and $- 1$ .

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

Step 14.3

Simplify the expression.

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Step 14.3.1

Move $- 1$ to the left of $1 - x$ .

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

Step 14.3.2

Rewrite $- 1 (1 - x)$ as $- (1 - x)$ .

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

Step 14.3.3

Move the negative in front of the fraction.

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