Calculus Examples

$y = \frac{e^{2 x}}{\sin (5 x)}$

Step 1

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = e^{2 x}$ and $g (x) = \sin (5 x)$ .

$\frac{\sin (5 x) \frac{d}{d x} [e^{2 x}] - e^{2 x} \frac{d}{d x} [\sin (5 x)]}{\sin^{2} (5 x)}$

Step 2

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

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Step 2.1

To apply the Chain Rule, set $u_{1}$ as $2 x$ .

$\frac{\sin (5 x) (\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d x} [2 x]) - e^{2 x} \frac{d}{d x} [\sin (5 x)]}{\sin^{2} (5 x)}$

Step 2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$\frac{\sin (5 x) (e^{u_{1}} \frac{d}{d x} [2 x]) - e^{2 x} \frac{d}{d x} [\sin (5 x)]}{\sin^{2} (5 x)}$

Step 2.3

Replace all occurrences of $u_{1}$ with $2 x$ .

$\frac{\sin (5 x) (e^{2 x} \frac{d}{d x} [2 x]) - e^{2 x} \frac{d}{d x} [\sin (5 x)]}{\sin^{2} (5 x)}$

Step 3

Differentiate.

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Step 3.1

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

$\frac{\sin (5 x) (e^{2 x} (2 \frac{d}{d x} [x])) - e^{2 x} \frac{d}{d x} [\sin (5 x)]}{\sin^{2} (5 x)}$

Step 3.2

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

$\frac{\sin (5 x) (e^{2 x} (2 \cdot 1)) - e^{2 x} \frac{d}{d x} [\sin (5 x)]}{\sin^{2} (5 x)}$

Step 3.3

Simplify the expression.

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Step 3.3.1

Multiply $2$ by $1$ .

$\frac{\sin (5 x) (e^{2 x} \cdot 2) - e^{2 x} \frac{d}{d x} [\sin (5 x)]}{\sin^{2} (5 x)}$

Step 3.3.2

Move $2$ to the left of $e^{2 x}$ .

$\frac{\sin (5 x) (2 \cdot e^{2 x}) - e^{2 x} \frac{d}{d x} [\sin (5 x)]}{\sin^{2} (5 x)}$

Step 4

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 $g (x) = 5 x$ .

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Step 4.1

To apply the Chain Rule, set $u_{2}$ as $5 x$ .

$\frac{\sin (5 x) (2 e^{2 x}) - e^{2 x} (\frac{d}{d u_{2}} [\sin (u_{2})] \frac{d}{d x} [5 x])}{\sin^{2} (5 x)}$

Step 4.2

The derivative of $\sin (u_{2})$ with respect to $u_{2}$ is $\cos (u_{2})$ .

$\frac{\sin (5 x) (2 e^{2 x}) - e^{2 x} (\cos (u_{2}) \frac{d}{d x} [5 x])}{\sin^{2} (5 x)}$

Step 4.3

Replace all occurrences of $u_{2}$ with $5 x$ .

$\frac{\sin (5 x) (2 e^{2 x}) - e^{2 x} (\cos (5 x) \frac{d}{d x} [5 x])}{\sin^{2} (5 x)}$

Step 5

Differentiate.

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Step 5.1

Since $5$ is constant with respect to $x$ , the derivative of $5 x$ with respect to $x$ is $5 \frac{d}{d x} [x]$ .

$\frac{\sin (5 x) (2 e^{2 x}) - e^{2 x} (\cos (5 x) (5 \frac{d}{d x} [x]))}{\sin^{2} (5 x)}$

Step 5.2

Multiply $5$ by $- 1$ .

$\frac{\sin (5 x) (2 e^{2 x}) - 5 e^{2 x} (\cos (5 x) (\frac{d}{d x} [x]))}{\sin^{2} (5 x)}$

Step 5.3

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

$\frac{\sin (5 x) (2 e^{2 x}) - 5 e^{2 x} (\cos (5 x) \cdot 1)}{\sin^{2} (5 x)}$

Step 5.4

Multiply $\cos (5 x)$ by $1$ .

$\frac{\sin (5 x) (2 e^{2 x}) - 5 e^{2 x} \cos (5 x)}{\sin^{2} (5 x)}$

Step 6

Simplify.

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Step 6.1

Simplify the numerator.

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Step 6.1.1

Rewrite using the commutative property of multiplication.

$\frac{2 \sin (5 x) e^{2 x} - 5 e^{2 x} \cos (5 x)}{\sin^{2} (5 x)}$

Step 6.1.2

Reorder factors in $2 \sin (5 x) e^{2 x} - 5 e^{2 x} \cos (5 x)$ .

$\frac{2 e^{2 x} \sin (5 x) - 5 e^{2 x} \cos (5 x)}{\sin^{2} (5 x)}$

Step 6.2

Factor $e^{2 x}$ out of $2 e^{2 x} \sin (5 x) - 5 e^{2 x} \cos (5 x)$ .

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Step 6.2.1

Factor $e^{2 x}$ out of $2 e^{2 x} \sin (5 x)$ .

$\frac{e^{2 x} (2 \sin (5 x)) - 5 e^{2 x} \cos (5 x)}{\sin^{2} (5 x)}$

Step 6.2.2

Factor $e^{2 x}$ out of $- 5 e^{2 x} \cos (5 x)$ .

$\frac{e^{2 x} (2 \sin (5 x)) + e^{2 x} (- 5 \cos (5 x))}{\sin^{2} (5 x)}$

Step 6.2.3

Factor $e^{2 x}$ out of $e^{2 x} (2 \sin (5 x)) + e^{2 x} (- 5 \cos (5 x))$ .

$\frac{e^{2 x} (2 \sin (5 x) - 5 \cos (5 x))}{\sin^{2} (5 x)}$