Calculus Examples

\cos (2 y)

$\cos (2 y)$

Step 1

Differentiate using the chain rule, which states that

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

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

f' (g (y)) g' (y)

$f' (g (y)) g' (y)$ where

f (y) = \cos (y)

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

g (y) = 2 y

$g (y) = 2 y$ .

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Step 1.1

To apply the Chain Rule, set

u

$u$ as

2 y

$2 y$ .

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

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

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 y} [2 y]

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

Step 1.3

Replace all occurrences of

u

$u$ with

2 y

$2 y$ .

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

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

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

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

Step 2

Differentiate.

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Step 2.1

Since

2

$2$ is constant with respect to

y

$y$ , the derivative of

2 y

$2 y$ with respect to

y

$y$ is

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

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

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

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

Step 2.2

Multiply

2

$2$ by

- 1

$- 1$ .

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

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

Step 2.3

Differentiate using the Power Rule which states that

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

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

n y^{n - 1}

$n y^{n - 1}$ where

n = 1

$n = 1$ .

- 2 \sin (2 y) \cdot 1

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

Step 2.4

Multiply

- 2

$- 2$ by

1

$1$ .

- 2 \sin (2 y)

$- 2 \sin (2 y)$

- 2 \sin (2 y)

$- 2 \sin (2 y)$

\cos 2 y

$\cos 2 y$

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