# Derivative of sin *x*

How to solve the derivative of sin *x* problems: formula, proof, examples, and their solutions.

## Formula

The derivative of sin *x* is cos *x*.

## Proof

Definition of a derivative function

Write the gray term

at the front part.

And combine the sin *x* terms.

Split the fraction into two parts.

Write the denominators *h*

under (sin *h*) and (cos *h* - 1).

Recall that as *h* → 0,

(sin *h*)/*h* → 1.

But (cos *h* - 1)/*h* is 0/0 form.

So multiply the conjugate of (cos *h* - 1), (cos *h* + 1)

to both of the numerator and the denominator.

Limits of trigonometric functions

sin^{2} *h* + cos^{2} *h* = 1

So cos^{2} *h* - 1 = -sin^{2} *h*.

Pythagorean identities

Change -sin^{2} *h* into sin *h*.

And write -sin *h*

on the numerator of 1/(cos *h* + 1).

As *h* → 0,

(sin *h*)/*h* → 1 (gray)

and -sin *h* → 0. (red)

So the limit is

cos *x*⋅1 + sin *x*⋅1⋅(-0/(1 + 1)),

which is cos *x*.

So [sin *x*]' = cos *x*.

## Example 1

*y* = (*x*)(sin *x*)

So *y*' is equal to,

the derivative of *x*, 1

times sin *x*

plus *x*,

times, the derivative of sin *x*, cos *x*.

Product rule in differentiation

Power rule in differentiation (Part 1)

## Example 2

*y* = sin (*x*^{2})

So *y*' is equal to,

the derivative of the outer part, cos (*x*^{2})

times, the derivative of the inner part, 2*x*.

Chain rule in differentiation

Power rule in differentiation (Part 1)

## Example 3

The given function is slightly different

from the previous example's function.*y* = (sin *x*)^{2}

So *y*' is equal to,

the derivative of the outer part, 2(sin *x*)

times, the derivative of the inner part, cos *x*.

Chain rule in differentiation

Power rule in differentiation (Part 1)