Calculus Examples

$f (x) = 9 \sin^{2} (x) - 9 \cos (x)$

Step 1

By the Sum Rule, the derivative of $9 \sin^{2} (x) - 9 \cos (x)$ with respect to $x$ is $\frac{d}{d x} [9 \sin^{2} (x)] + \frac{d}{d x} [- 9 \cos (x)]$ .

$\frac{d}{d x} [9 \sin^{2} (x)] + \frac{d}{d x} [- 9 \cos (x)]$

Step 2

Evaluate $\frac{d}{d x} [9 \sin^{2} (x)]$ .

Tap for more steps...

Step 2.1

Since $9$ is constant with respect to $x$ , the derivative of $9 \sin^{2} (x)$ with respect to $x$ is $9 \frac{d}{d x} [\sin^{2} (x)]$ .

$9 \frac{d}{d x} [\sin^{2} (x)] + \frac{d}{d x} [- 9 \cos (x)]$

Step 2.2

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

Tap for more steps...

Step 2.2.1

To apply the Chain Rule, set $u$ as $\sin (x)$ .

$9 (\frac{d}{d u} [u^{2}] \frac{d}{d x} [\sin (x)]) + \frac{d}{d x} [- 9 \cos (x)]$

Step 2.2.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 2$ .

$9 (2 u \frac{d}{d x} [\sin (x)]) + \frac{d}{d x} [- 9 \cos (x)]$

Step 2.2.3

Replace all occurrences of $u$ with $\sin (x)$ .

$9 (2 \sin (x) \frac{d}{d x} [\sin (x)]) + \frac{d}{d x} [- 9 \cos (x)]$

Step 2.3

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

$9 (2 \sin (x) \cos (x)) + \frac{d}{d x} [- 9 \cos (x)]$

Step 2.4

Multiply $2$ by $9$ .

$18 \sin (x) \cos (x) + \frac{d}{d x} [- 9 \cos (x)]$

Step 3

Evaluate $\frac{d}{d x} [- 9 \cos (x)]$ .

Tap for more steps...

Step 3.1

Since $- 9$ is constant with respect to $x$ , the derivative of $- 9 \cos (x)$ with respect to $x$ is $- 9 \frac{d}{d x} [\cos (x)]$ .

$18 \sin (x) \cos (x) - 9 \frac{d}{d x} [\cos (x)]$

Step 3.2

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

$18 \sin (x) \cos (x) - 9 (- \sin (x))$

Step 3.3

Multiply $- 1$ by $- 9$ .

$18 \sin (x) \cos (x) + 9 \sin (x)$

Step 4

Reorder terms.

$18 \cos (x) \sin (x) + 9 \sin (x)$