Precalculus Examples

$f (x) = {(2 x^{3} + 3)}^{4}$

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 $g (x) = 2 x^{3} + 3$ .

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Step 1.1

To apply the Chain Rule, set $u$ as $2 x^{3} + 3$ .

$\frac{d}{d u} [u^{4}] \frac{d}{d x} [2 x^{3} + 3]$

Step 1.2

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

$4 u^{3} \frac{d}{d x} [2 x^{3} + 3]$

Step 1.3

Replace all occurrences of $u$ with $2 x^{3} + 3$ .

$4 {(2 x^{3} + 3)}^{3} \frac{d}{d x} [2 x^{3} + 3]$

Step 2

Differentiate.

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Step 2.1

By the Sum Rule, the derivative of $2 x^{3} + 3$ with respect to $x$ is $\frac{d}{d x} [2 x^{3}] + \frac{d}{d x} [3]$ .

$4 {(2 x^{3} + 3)}^{3} (\frac{d}{d x} [2 x^{3}] + \frac{d}{d x} [3])$

Step 2.2

Since $2$ is constant with respect to $x$ , the derivative of $2 x^{3}$ with respect to $x$ is $2 \frac{d}{d x} [x^{3}]$ .

$4 {(2 x^{3} + 3)}^{3} (2 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [3])$

Step 2.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$4 {(2 x^{3} + 3)}^{3} (2 (3 x^{2}) + \frac{d}{d x} [3])$

Step 2.4

Multiply $3$ by $2$ .

$4 {(2 x^{3} + 3)}^{3} (6 x^{2} + \frac{d}{d x} [3])$

Step 2.5

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

$4 {(2 x^{3} + 3)}^{3} (6 x^{2} + 0)$

Step 2.6

Simplify the expression.

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Step 2.6.1

Add $6 x^{2}$ and $0$ .

$4 {(2 x^{3} + 3)}^{3} (6 x^{2})$

Step 2.6.2

Multiply $6$ by $4$ .

$24 {(2 x^{3} + 3)}^{3} x^{2}$

Step 2.6.3

Reorder the factors of $24 {(2 x^{3} + 3)}^{3} x^{2}$ .

$24 x^{2} {(2 x^{3} + 3)}^{3}$