Calculus Examples

$\frac{1 + \sin (x)}{1 - \sin (x)}$

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 $f (x) = 1 + \sin (x)$ and $g (x) = 1 - \sin (x)$ .

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

Step 2

Differentiate.

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Step 2.1

By the Sum Rule, the derivative of $1 + \sin (x)$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [\sin (x)]$ .

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

Step 2.2

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$\frac{(1 - \sin (x)) (0 + \frac{d}{d x} [\sin (x)]) - (1 + \sin (x)) \frac{d}{d x} [1 - \sin (x)]}{{(1 - \sin (x))}^{2}}$

Step 2.3

Add $0$ and $\frac{d}{d x} [\sin (x)]$ .

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

Step 3

The derivative of $\sin (x)$ with respect to $x$ is $\cos (x)$ .

$\frac{(1 - \sin (x)) \cos (x) - (1 + \sin (x)) \frac{d}{d x} [1 - \sin (x)]}{{(1 - \sin (x))}^{2}}$

Step 4

Differentiate.

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Step 4.1

By the Sum Rule, the derivative of $1 - \sin (x)$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- \sin (x)]$ .

$\frac{(1 - \sin (x)) \cos (x) - (1 + \sin (x)) (\frac{d}{d x} [1] + \frac{d}{d x} [- \sin (x)])}{{(1 - \sin (x))}^{2}}$

Step 4.2

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$\frac{(1 - \sin (x)) \cos (x) - (1 + \sin (x)) (0 + \frac{d}{d x} [- \sin (x)])}{{(1 - \sin (x))}^{2}}$

Step 4.3

Add $0$ and $\frac{d}{d x} [- \sin (x)]$ .

$\frac{(1 - \sin (x)) \cos (x) - (1 + \sin (x)) \frac{d}{d x} [- \sin (x)]}{{(1 - \sin (x))}^{2}}$

Step 4.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- \sin (x)$ with respect to $x$ is $- \frac{d}{d x} [\sin (x)]$ .

$\frac{(1 - \sin (x)) \cos (x) - (1 + \sin (x)) (- \frac{d}{d x} [\sin (x)])}{{(1 - \sin (x))}^{2}}$

Step 4.5

Multiply.

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Step 4.5.1

Multiply $- 1$ by $- 1$ .

$\frac{(1 - \sin (x)) \cos (x) + 1 (1 + \sin (x)) \frac{d}{d x} [\sin (x)]}{{(1 - \sin (x))}^{2}}$

Step 4.5.2

Multiply $1 + \sin (x)$ by $1$ .

$\frac{(1 - \sin (x)) \cos (x) + (1 + \sin (x)) \frac{d}{d x} [\sin (x)]}{{(1 - \sin (x))}^{2}}$

Step 5

The derivative of $\sin (x)$ with respect to $x$ is $\cos (x)$ .

$\frac{(1 - \sin (x)) \cos (x) + (1 + \sin (x)) \cos (x)}{{(1 - \sin (x))}^{2}}$

Step 6

Simplify.

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Step 6.1

Apply the distributive property.

$\frac{1 \cos (x) - \sin (x) \cos (x) + (1 + \sin (x)) \cos (x)}{{(1 - \sin (x))}^{2}}$

Step 6.2

Apply the distributive property.

$\frac{1 \cos (x) - \sin (x) \cos (x) + 1 \cos (x) + \sin (x) \cos (x)}{{(1 - \sin (x))}^{2}}$

Step 6.3

Simplify the numerator.

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Step 6.3.1

Combine the opposite terms in $1 \cos (x) - \sin (x) \cos (x) + 1 \cos (x) + \sin (x) \cos (x)$ .

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Step 6.3.1.1

Add $- \sin (x) \cos (x)$ and $\sin (x) \cos (x)$ .

$\frac{1 \cos (x) + 0 + 1 \cos (x)}{{(1 - \sin (x))}^{2}}$

Step 6.3.1.2

Add $1 \cos (x)$ and $0$ .

$\frac{1 \cos (x) + 1 \cos (x)}{{(1 - \sin (x))}^{2}}$

Step 6.3.2

Simplify each term.

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Step 6.3.2.1

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

$\frac{\cos (x) + 1 \cos (x)}{{(1 - \sin (x))}^{2}}$

Step 6.3.2.2

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

$\frac{\cos (x) + \cos (x)}{{(1 - \sin (x))}^{2}}$

Step 6.3.3

Add $\cos (x)$ and $\cos (x)$ .

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