Calculus Examples

f (x) = cos (2 x)

$f (x) = \cos (2 x)$

Step 1

Differentiate using the chain rule, which states that

\frac{d}{d x} [f (g (x))]

$\frac{d}{d x} [f (g (x))]$ is

f' (g (x)) g' (x)

$f' (g (x)) g' (x)$ where

f (x) = cos (x)

$f (x) = \cos (x)$ and

g (x) = 2 x

$g (x) = 2 x$ .

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Step 1.1

To apply the Chain Rule, set

u

$u$ as

2 x

$2 x$ .

\frac{d}{d u} [cos (u)] \frac{d}{d x} [2 x]

$\frac{d}{d u} [\cos (u)] \frac{d}{d x} [2 x]$

Step 1.2

The derivative of

cos (u)

$\cos (u)$ with respect to

u

$u$ is

- sin (u)

$- \sin (u)$ .

- sin (u) \frac{d}{d x} [2 x]

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

Step 1.3

Replace all occurrences of

u

$u$ with

2 x

$2 x$ .

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

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

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

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

Step 2

Differentiate.

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Step 2.1

Since

2

$2$ is constant with respect to

x

$x$ , the derivative of

2 x

$2 x$ with respect to

x

$x$ is

2 \frac{d}{d x} [x]

$2 \frac{d}{d x} [x]$ .

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

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

Step 2.2

Multiply

2

$2$ by

- 1

$- 1$ .

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

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

Step 2.3

Differentiate using the Power Rule which states that

\frac{d}{d x} [x^{n}]

$\frac{d}{d x} [x^{n}]$ is

n x^{n - 1}

$n x^{n - 1}$ where

n = 1

$n = 1$ .

- 2 sin (2 x) \cdot 1

$- 2 \sin (2 x) \cdot 1$

Step 2.4

Multiply

- 2

$- 2$ by

1

$1$ .

- 2 sin (2 x)

$- 2 \sin (2 x)$

- 2 sin (2 x)

$- 2 \sin (2 x)$