Calculus Examples

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

Step 1

Rewrite $\frac{1}{e^{x} + 1}$ as ${(e^{x} + 1)}^{- 1}$ .

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

Step 2

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

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Step 2.1

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

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

Step 2.2

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

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

Step 2.3

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

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

Step 3

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

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

Step 4

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

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

Step 5

Differentiate using the Constant Rule.

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Step 5.1

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

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

Step 5.2

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

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

Step 6

Simplify.

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Step 6.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 6.2

Combine $e^{x}$ and $\frac{1}{{(e^{x} + 1)}^{2}}$ .

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