Calculus Examples

$\frac{1}{{(x + 7)}^{2}}$

Step 1

Apply basic rules of exponents.

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Step 1.1

Rewrite $\frac{1}{{(x + 7)}^{2}}$ as ${({(x + 7)}^{2})}^{- 1}$ .

$\frac{d}{d x} [{({(x + 7)}^{2})}^{- 1}]$

Step 1.2

Multiply the exponents in ${({(x + 7)}^{2})}^{- 1}$ .

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Step 1.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{d}{d x} [{(x + 7)}^{2 \cdot - 1}]$

Step 1.2.2

Multiply $2$ by $- 1$ .

$\frac{d}{d x} [{(x + 7)}^{- 2}]$

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 2}$ and $g (x) = x + 7$ .

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Step 2.1

To apply the Chain Rule, set $u$ as $x + 7$ .

$\frac{d}{d u} [u^{- 2}] \frac{d}{d x} [x + 7]$

Step 2.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = - 2$ .

$- 2 u^{- 3} \frac{d}{d x} [x + 7]$

Step 2.3

Replace all occurrences of $u$ with $x + 7$ .

$- 2 {(x + 7)}^{- 3} \frac{d}{d x} [x + 7]$

Step 3

Differentiate.

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Step 3.1

By the Sum Rule, the derivative of $x + 7$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [7]$ .

$- 2 {(x + 7)}^{- 3} (\frac{d}{d x} [x] + \frac{d}{d x} [7])$

Step 3.2

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

$- 2 {(x + 7)}^{- 3} (1 + \frac{d}{d x} [7])$

Step 3.3

Since $7$ is constant with respect to $x$ , the derivative of $7$ with respect to $x$ is $0$ .

$- 2 {(x + 7)}^{- 3} (1 + 0)$

Step 3.4

Simplify the expression.

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Step 3.4.1

Add $1$ and $0$ .

$- 2 {(x + 7)}^{- 3} \cdot 1$

Step 3.4.2

Multiply $- 2$ by $1$ .

$- 2 {(x + 7)}^{- 3}$

Step 4

Simplify.

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Step 4.1

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

$- 2 \frac{1}{{(x + 7)}^{3}}$

Step 4.2

Combine terms.

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Step 4.2.1

Combine $- 2$ and $\frac{1}{{(x + 7)}^{3}}$ .

$\frac{- 2}{{(x + 7)}^{3}}$

Step 4.2.2

Move the negative in front of the fraction.

$- \frac{2}{{(x + 7)}^{3}}$