Calculus Examples

${(8 x - 9)}^{- 4}$

Step 1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 4}$ and $g (x) = 8 x - 9$ .

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Step 1.1

To apply the Chain Rule, set $u$ as $8 x - 9$ .

$\frac{d}{d u} [u^{- 4}] \frac{d}{d x} [8 x - 9]$

Step 1.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = - 4$ .

$- 4 u^{- 5} \frac{d}{d x} [8 x - 9]$

Step 1.3

Replace all occurrences of $u$ with $8 x - 9$ .

$- 4 {(8 x - 9)}^{- 5} \frac{d}{d x} [8 x - 9]$

Step 2

Differentiate.

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Step 2.1

By the Sum Rule, the derivative of $8 x - 9$ with respect to $x$ is $\frac{d}{d x} [8 x] + \frac{d}{d x} [- 9]$ .

$- 4 {(8 x - 9)}^{- 5} (\frac{d}{d x} [8 x] + \frac{d}{d x} [- 9])$

Step 2.2

Since $8$ is constant with respect to $x$ , the derivative of $8 x$ with respect to $x$ is $8 \frac{d}{d x} [x]$ .

$- 4 {(8 x - 9)}^{- 5} (8 \frac{d}{d x} [x] + \frac{d}{d x} [- 9])$

Step 2.3

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

$- 4 {(8 x - 9)}^{- 5} (8 \cdot 1 + \frac{d}{d x} [- 9])$

Step 2.4

Multiply $8$ by $1$ .

$- 4 {(8 x - 9)}^{- 5} (8 + \frac{d}{d x} [- 9])$

Step 2.5

Since $- 9$ is constant with respect to $x$ , the derivative of $- 9$ with respect to $x$ is $0$ .

$- 4 {(8 x - 9)}^{- 5} (8 + 0)$

Step 2.6

Simplify the expression.

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Step 2.6.1

Add $8$ and $0$ .

$- 4 {(8 x - 9)}^{- 5} \cdot 8$

Step 2.6.2

Multiply $8$ by $- 4$ .

$- 32 {(8 x - 9)}^{- 5}$

Step 3

Simplify.

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Step 3.1

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

$- 32 \frac{1}{{(8 x - 9)}^{5}}$

Step 3.2

Combine $- 32$ and $\frac{1}{{(8 x - 9)}^{5}}$ .

$\frac{- 32}{{(8 x - 9)}^{5}}$

Step 3.3

Move the negative in front of the fraction.

$- \frac{32}{{(8 x - 9)}^{5}}$