Calculus Examples

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

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

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

Step 2

Differentiate.

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

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

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

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

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

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

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

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

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

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

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

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

Multiply.

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Multiply $- 1$ by $- 1$ .

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

Multiply $x$ by $1$ .

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

Step 3

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

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

Step 4

Reorder terms.

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