Calculus Examples

${(2 x^{7} - 4 x)}^{8}$

Step 1

Differentiate using the chain rule, which states that

$\frac{d}{d x} [f (g (x))]$ is

$f' (g (x)) g' (x)$ where

$f (x) = x^{8}$ and

$g (x) = 2 x^{7} - 4 x$ .

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Step 1.1

To apply the Chain Rule, set

$u$ as

$2 x^{7} - 4 x$ .

$\frac{d}{d u} [u^{8}] \frac{d}{d x} [2 x^{7} - 4 x]$

Step 1.2

Differentiate using the Power Rule which states that

$\frac{d}{d u} [u^{n}]$ is

$n u^{n - 1}$ where

$n = 8$ .

$8 u^{7} \frac{d}{d x} [2 x^{7} - 4 x]$

Step 1.3

Replace all occurrences of

$u$ with

$2 x^{7} - 4 x$ .

$8 {(2 x^{7} - 4 x)}^{7} \frac{d}{d x} [2 x^{7} - 4 x]$

Step 2

By the Sum Rule, the derivative of

$2 x^{7} - 4 x$ with respect to

$x$ is

$\frac{d}{d x} [2 x^{7}] + \frac{d}{d x} [- 4 x]$ .

$8 {(2 x^{7} - 4 x)}^{7} (\frac{d}{d x} [2 x^{7}] + \frac{d}{d x} [- 4 x])$

Step 3

Evaluate

$\frac{d}{d x} [2 x^{7}]$ .

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Step 3.1

Since

$2$ is constant with respect to

$x$ , the derivative of

$2 x^{7}$ with respect to

$x$ is

$2 \frac{d}{d x} [x^{7}]$ .

$8 {(2 x^{7} - 4 x)}^{7} (2 \frac{d}{d x} [x^{7}] + \frac{d}{d x} [- 4 x])$

Step 3.2

Differentiate using the Power Rule which states that

$\frac{d}{d x} [x^{n}]$ is

$n x^{n - 1}$ where

$n = 7$ .

$8 {(2 x^{7} - 4 x)}^{7} (2 (7 x^{6}) + \frac{d}{d x} [- 4 x])$

Step 3.3

Multiply

$7$ by

$2$ .

$8 {(2 x^{7} - 4 x)}^{7} (14 x^{6} + \frac{d}{d x} [- 4 x])$

Step 4

Evaluate

$\frac{d}{d x} [- 4 x]$ .

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Step 4.1

Since

$- 4$ is constant with respect to

$x$ , the derivative of

$- 4 x$ with respect to

$x$ is

$- 4 \frac{d}{d x} [x]$ .

$8 {(2 x^{7} - 4 x)}^{7} (14 x^{6} - 4 \frac{d}{d x} [x])$

Step 4.2

Differentiate using the Power Rule which states that

$\frac{d}{d x} [x^{n}]$ is

$n x^{n - 1}$ where

$n = 1$ .

$8 {(2 x^{7} - 4 x)}^{7} (14 x^{6} - 4 \cdot 1)$

Step 4.3

Multiply

$- 4$ by

$1$ .

$8 {(2 x^{7} - 4 x)}^{7} (14 x^{6} - 4)$

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