Calculus Examples

$\ln (x + 1)$

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

$g (x) = x + 1$ .

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Step 1.1

To apply the Chain Rule, set

$u$ as

$x + 1$ .

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

Step 1.2

The derivative of

$\ln (u)$ with respect to

$u$ is

$\frac{1}{u}$ .

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

Step 1.3

Replace all occurrences of

$u$ with

$x + 1$ .

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

Step 2

Differentiate.

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Step 2.1

By the Sum Rule, the derivative of

$x + 1$ with respect to

$x$ is

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

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

Step 2.2

Differentiate using the Power Rule which states that

$\frac{d}{d x} [x^{n}]$ is

$n x^{n - 1}$ where

$n = 1$ .

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

Step 2.3

Since

$1$ is constant with respect to

$x$ , the derivative of

$1$ with respect to

$x$ is

$0$ .

$\frac{1}{x + 1} (1 + 0)$

Step 2.4

Simplify the expression.

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Step 2.4.1

Add

$1$ and

$0$ .

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

Step 2.4.2

Multiply

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

$1$ .

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