Calculus Examples

${(e^{3 x} + \sin (2 x))}^{4}$

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

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Step 1.1

To apply the Chain Rule, set $u_{1}$ as $e^{3 x} + \sin (2 x)$ .

$\frac{d}{d u_{1}} [{u_{1}}^{4}] \frac{d}{d x} [e^{3 x} + \sin (2 x)]$

Step 1.2

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

$4 {u_{1}}^{3} \frac{d}{d x} [e^{3 x} + \sin (2 x)]$

Step 1.3

Replace all occurrences of $u_{1}$ with $e^{3 x} + \sin (2 x)$ .

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

Step 2

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

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

Step 3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = 3 x$ .

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Step 3.1

To apply the Chain Rule, set $u_{2}$ as $3 x$ .

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

Step 3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

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

Step 3.3

Replace all occurrences of $u_{2}$ with $3 x$ .

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

Step 4

Differentiate.

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Step 4.1

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

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

Step 4.2

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

$4 {(e^{3 x} + \sin (2 x))}^{3} (e^{3 x} (3 \cdot 1) + \frac{d}{d x} [\sin (2 x)])$

Step 4.3

Simplify the expression.

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Step 4.3.1

Multiply $3$ by $1$ .

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

Step 4.3.2

Move $3$ to the left of $e^{3 x}$ .

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

Step 5

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

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Step 5.1

To apply the Chain Rule, set $u_{3}$ as $2 x$ .

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

Step 5.2

The derivative of $\sin (u_{3})$ with respect to $u_{3}$ is $\cos (u_{3})$ .

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

Step 5.3

Replace all occurrences of $u_{3}$ with $2 x$ .

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

Step 6

Differentiate.

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Step 6.1

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

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

Step 6.2

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

$4 {(e^{3 x} + \sin (2 x))}^{3} (3 e^{3 x} + \cos (2 x) (2 \cdot 1))$

Step 6.3

Simplify the expression.

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Step 6.3.1

Multiply $2$ by $1$ .

$4 {(e^{3 x} + \sin (2 x))}^{3} (3 e^{3 x} + \cos (2 x) \cdot 2)$

Step 6.3.2

Move $2$ to the left of $\cos (2 x)$ .

$4 {(e^{3 x} + \sin (2 x))}^{3} (3 e^{3 x} + 2 \cos (2 x))$