Calculus Examples

$\ln (2 e^{x} - 1)$

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 $g (x) = 2 e^{x} - 1$ .

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Step 1.1

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

$\frac{d}{d u} [\ln (u)] \frac{d}{d x} [2 e^{x} - 1]$

Step 1.2

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

$\frac{1}{u} \frac{d}{d x} [2 e^{x} - 1]$

Step 1.3

Replace all occurrences of $u$ with $2 e^{x} - 1$ .

$\frac{1}{2 e^{x} - 1} \frac{d}{d x} [2 e^{x} - 1]$

Step 2

Differentiate.

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Step 2.1

By the Sum Rule, the derivative of $2 e^{x} - 1$ with respect to $x$ is $\frac{d}{d x} [2 e^{x}] + \frac{d}{d x} [- 1]$ .

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

Step 2.2

Since $2$ is constant with respect to $x$ , the derivative of $2 e^{x}$ with respect to $x$ is $2 \frac{d}{d x} [e^{x}]$ .

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

Step 3

Differentiate using the Exponential Rule which states that $\frac{d}{d x} [a^{x}]$ is $a^{x} \ln (a)$ where $a$ = $e$ .

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

Step 4

Differentiate using the Constant Rule.

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Step 4.1

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

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

Step 4.2

Combine fractions.

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Step 4.2.1

Add $2 e^{x}$ and $0$ .

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

Step 4.2.2

Combine $2$ and $\frac{1}{2 e^{x} - 1}$ .

$\frac{2}{2 e^{x} - 1} e^{x}$

Step 4.2.3

Combine $\frac{2}{2 e^{x} - 1}$ and $e^{x}$ .

$\frac{2 e^{x}}{2 e^{x} - 1}$