Calculus Examples

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

Step 1

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = \ln (t)$ and $g (t) = 2 e^{t} - 1$ .

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Step 1.1

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

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

Step 1.2

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

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

Step 1.3

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

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

Step 2

Differentiate.

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Step 2.1

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

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

Step 2.2

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

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

Step 3

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

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

Step 4

Differentiate using the Constant Rule.

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Step 4.1

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

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

Step 4.2

Combine fractions.

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Step 4.2.1

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

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

Step 4.2.2

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

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

Step 4.2.3

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

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