# Derivative of sin^2x by First Principle | sin^2x Derivative

The derivative of sin^2x by first principle is equal to 2sinx cosx=sin2x. Sin square x derivative is denoted by d/dx (sin2x) and its formula is given as follows:

$\dfrac{d}{dx}$(sin2x) = 2sinx cosx = sin2x.

In this post, we will find the derivative of sine square x.

## Derivative of sin square x by First Principle

The first principle of derivatives says that the derivative of a function f(x) is given by the following limit:

$\dfrac{d}{dx}(f(x))$ = limh→0 $\dfrac{f(x+h)-f(x)}{h}$.

To find the derivative of sin2x, we put f(x)=sin2x in the formula above. So the derivative of sin square x will be equal to

$\dfrac{d}{dx}(\sin^2x)$ = limh→0 $\dfrac{\sin^2(x+h)-\sin^2x}{h}$

= limh→0 $\dfrac{( \sin x \cos h +\cos x \sin h)^2 – \sin^2x}{h}$  by the formula sin(x+h)= sinx cosh +cosx sinh.

= limh→0 $(\sin^2 x \cos^2 h +2\sin x \cos h\cos x \sin h$ $+\cos^2x\sin^2h – \sin^2x)$/h,  as we know that $(a+b)^2$ $=a^2+2ab+b^2$

= limh→0 $\Big[\dfrac{\sin^2 x (\cos^2 h-1)}{h}$ $+ \dfrac{2\sin x \cos h\cos x \sin h}{h}$ $+\dfrac{\cos^2x\sin^2h}{h} \Big]$

Let us now apply the Sum rule of limits.

= limh→0 $\dfrac{\sin^2 x (\cos^2 h-1)}{h}$ + limh→0 $\dfrac{2\sin x \cos h\cos x \sin h}{h}$ + limh→0 $\dfrac{\cos^2x\sin^2h}{h}$

= sin2x limh→0 $\dfrac{(\cos h-1)(\cos h+1)}{h}$ + 2 limh→0 $\left(\sin x \cos h\cos x\dfrac{\sin h}{h} \right)$ + limh→0 $\left( \cos^2 x \sin h \dfrac{\sin h}{h} \right)$

Now, as the limit limh→0(sinh/h) is 1, the above will be

= sin2x (cos 0+1) limh→0 $\dfrac{\cos h-1}{h}$ + 2 sinx cos0 cosx ⋅ 1+ cos2x sin0⋅1

= 2 sin2x limh→0 $\dfrac{\cos h-1}{h}$ + 2 sinx cosx + 0

By the L’Hospital rule, the limit limh→0$\dfrac{\cos h-1}{h}$ = limh→0 $\dfrac{-\sin h}{1}$ =0.

So the above quantity is equal to

2 sin2x ⋅ 0 + 2 sinx cosx + 0

= 0 + 2 sinx cos x + 0

= 2 sin x cos x.

So the der

So the derivative of sin2x by the first principle is equal to 2sin x cos x which is further equal to sin2x.

You Can also Read These Derivatives:

Derivative of cos2x by first principle

Derivative of $\sqrt{\sin x}$ by first principle

Derivative of $\sqrt{\cos x}$ by first principle

Derivative of mod sinx

## FAQs

### Q1: What is the derivative of sin^2x (sine square x)?

Answer: The derivative of sin^2x (sin square x) is equal to 2sinx cosx.