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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