# Derivative of sin x: Proof by Quotient, Chain & First Principle

In this article, you will learn what is the derivative of sin x as well as prove the derivative of sin x by quotient rule, first principal rule, and chain rule.

I have already discussed the derivative of cos x, tan x, and sec x in a previous article you can check from here.

So, without wasting time let's get started.

## What is sin x?

Sin x is a trigonometric function that is reciprocal of cosec x.

## Derivative of sin x

The derivative of sin x is equal to cos x.

We can prove the derivative of sin x in three ways first by using the quotient rule and second by using the first principle rule and the last chain rule.

## Derivative of sin x Proof by Quotient Rule

The formula of the quotient rule is,

dy/dx = {v (du/dx) - u (dv/dx)}/v²

Where,

dy/dx = derivative of y with respect to x

v = variable v

du/dx = derivative of u with respect to x

u = variable u

dv/dx = derivative of v with respect to x

v = variable v

Let us,

y = sin x

As we know,

sin x  = 1/cosec x

So,

We can write as,

y  = 1/cosec x.

Where,

u = 1

v = cosec x

Now putting these values on the quotient rule formula, we will get

dy/dx = [cosec x d/dx(1) - 1 d/dx(cosec x)] / (cosec x)²

Since,

d/dx (cosec x) = -cosec x.cot x

d/dx (1) = 0

So,

dy/dx = {0 - (-cosec x.cot x  )}/ (cosec x)²

dy/dx = (cosec x.cot x )/(cosec x)²

dy/dx = (cot x)/(cosec x)

Since,

cot x = cos x/sin x

cosec x = 1/sin x

So,

dy/dx = (cos x/sin x)/(1/sin x)

dy/dx = cos x

d/dx (sin x) = cos x

Thus, we proved the derivative of sin x will be equal to cos x using the quotient rule method.

## Derivative of sin x Proof by First Principle Rule

According to the first principle rule, the derivative limit of a function can be determined by computing the formula:

For a differentiable function y = f (x)

We define its derivative w.r.t  x as :

dy/dx = f ' (x) = limₕ→₀ [f(x+h) - f(x)]/h

f'(x) = limₕ→₀ [f(x+h) - f(x)]/h

This limit is used to represent the instantaneous rate of change of the function f(x).

Let,

f (x) = sin x

So,

f(x + h) = sin (x + h)

Putting these values on the above first principle rules equation.

f' (x) = limₕ→₀ [sin (x + h) - sin x]/h

As we know,

sin C - sin D = 2 cos{(C + D)/2}.sin{(C - D)/2}

So,

f'(x) = limₕ→₀ [{2cos{(x + h + x)/2}.sin{(x + h - x)/2}] / h

= limₕ→₀ [2 cos{(2x + h)/2}.sin (h/2)]/h

= limₕ→₀ [cos{(2x + h)/2] · limₕ→₀ {sin (h/2)}/(h/2)]

As h →0,  and (h/2)→0

So,

f'(x) = limₕ→₀[cos{(2x + h + x)/2} ·lim₍ₕ/₂₎→₀ {sin (h/2)}/ (h/2)]

Now, using limit formulas,

lim ₓ→₀(sin x/x) = 1

So,

f'(x) = [cos{(2x + 0)/2}·(1) = cos (2x/2) = cos x

f'(x) = cos x

Thus, we proved the derivative of sin x will be equal to cos x using the first principle rule method.

## Derivative of sin x Proof by Chain Rule

Let us,

y = sin x

As we know,

sin x = cos (π/2 - x)

So,

y = cos (π/2 - x)

By using the chain rule,

The formula of the chain rule is,

dy/dx = (dy/du) × (du/dx)

Where,

dy/dx = derivative of y with respect to x

dy/du = derivative of y with respect to u

du/dx = derivative of u with respect to x

After putting these values we can find,

dy/dx = d/dx {cos (π/2 - x)}

Since,

d/dx(cos x) = -sin x

Hence,

dy/dx = -sin(π/2 - x) · d/dx (π/2 - x)

dy/dx = - sin(π/2 - x).(-1)

dy/dx = sin(π/2 - x)

Since,

sin(π/2 - x) = cos x

Hence,

d/dx (sin x) = cos x

Thus, we proved the derivative of sin x will be equal to the cos x using the chain rule method.

So friends here I discussed all aspects related to the derivative of sin x

I hope you enjoy this topic If you have any doubts then you can ask me through comments or direct mail. I will definitely reply to you.

Thank You.