Calculus Examples

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

Step 1

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

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

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 $g (x) = x + 1$ .

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Step 2.1

To apply the Chain Rule, set $u$ as $x + 1$ .

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

Step 2.2

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

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

Step 2.3

Replace all occurrences of $u$ with $x + 1$ .

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

Step 3

Differentiate.

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

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

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

Step 3.4

Simplify the expression.

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Step 3.4.1

Add $1$ and $0$ .

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

Step 3.4.2

Multiply $- 1$ by $1$ .

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

Step 4

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

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