Calculus Examples

$\csc^{2} (e^{4 x})$

Step 1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = \csc (e^{4 x})$ .

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Step 1.1

To apply the Chain Rule, set $u_{1}$ as $\csc (e^{4 x})$ .

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

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 $n = 2$ .

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

Step 1.3

Replace all occurrences of $u_{1}$ with $\csc (e^{4 x})$ .

$2 \csc (e^{4 x}) \frac{d}{d x} [\csc (e^{4 x})]$

Step 2

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

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Step 2.1

To apply the Chain Rule, set $u_{2}$ as $e^{4 x}$ .

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

Step 2.2

The derivative of $\csc (u_{2})$ with respect to $u_{2}$ is $- \csc (u_{2}) \cot (u_{2})$ .

$2 \csc (e^{4 x}) (- \csc (u_{2}) \cot (u_{2}) \frac{d}{d x} [e^{4 x}])$

Step 2.3

Replace all occurrences of $u_{2}$ with $e^{4 x}$ .

$2 \csc (e^{4 x}) (- \csc (e^{4 x}) \cot (e^{4 x}) \frac{d}{d x} [e^{4 x}])$

Step 3

Multiply $- 1$ by $2$ .

$- 2 \csc (e^{4 x}) (\csc (e^{4 x}) \cot (e^{4 x}) \frac{d}{d x} [e^{4 x}])$

Step 4

Raise $\csc (e^{4 x})$ to the power of $1$ .

$- 2 (\csc^{1} (e^{4 x}) \csc (e^{4 x})) (\cot (e^{4 x}) \frac{d}{d x} [e^{4 x}])$

Step 5

Raise $\csc (e^{4 x})$ to the power of $1$ .

$- 2 (\csc^{1} (e^{4 x}) \csc^{1} (e^{4 x})) (\cot (e^{4 x}) \frac{d}{d x} [e^{4 x}])$

Step 6

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$- 2 {\csc (e^{4 x})}^{1 + 1} (\cot (e^{4 x}) \frac{d}{d x} [e^{4 x}])$

Step 7

Add $1$ and $1$ .

$- 2 \csc^{2} (e^{4 x}) (\cot (e^{4 x}) \frac{d}{d x} [e^{4 x}])$

Step 8

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 $g (x) = 4 x$ .

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Step 8.1

To apply the Chain Rule, set $u_{3}$ as $4 x$ .

$- 2 \csc^{2} (e^{4 x}) \cot (e^{4 x}) (\frac{d}{d u_{3}} [e^{u_{3}}] \frac{d}{d x} [4 x])$

Step 8.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{3}} [a^{u_{3}}]$ is $a^{u_{3}} \ln (a)$ where $a$ = $e$ .

$- 2 \csc^{2} (e^{4 x}) \cot (e^{4 x}) (e^{u_{3}} \frac{d}{d x} [4 x])$

Step 8.3

Replace all occurrences of $u_{3}$ with $4 x$ .

$- 2 \csc^{2} (e^{4 x}) \cot (e^{4 x}) (e^{4 x} \frac{d}{d x} [4 x])$

Step 9

Differentiate.

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Step 9.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$- 2 \csc^{2} (e^{4 x}) \cot (e^{4 x}) (e^{4 x} (4 \frac{d}{d x} [x]))$

Step 9.2

Multiply $4$ by $- 2$ .

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

Step 9.3

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

$- 8 \csc^{2} (e^{4 x}) \cot (e^{4 x}) (e^{4 x} \cdot 1)$

Step 9.4

Simplify the expression.

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Step 9.4.1

Multiply $e^{4 x}$ by $1$ .

$- 8 \csc^{2} (e^{4 x}) \cot (e^{4 x}) e^{4 x}$

Step 9.4.2

Reorder the factors of $- 8 \csc^{2} (e^{4 x}) \cot (e^{4 x}) e^{4 x}$ .

$- 8 e^{4 x} \csc^{2} (e^{4 x}) \cot (e^{4 x})$