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

The function sin square x is the product of two sine functions. The derivative of sin square x is sin 2x. In this post, we will find the derivative of sine square x.

## Derivative of sin^2 x by First Principle

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

$\dfrac{d}{dx}(f(x))$ $=\lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$.

In the above formula, we put $f(x)=\sin^2x$. So the derivative of \sin square x is equal \to

$\dfrac{d}{dx}(\sin^2x)$ $= \lim\limits_{h \to 0} \dfrac{\sin^2(x+h)-\sin^2x}{h}$

$= \lim\limits_{h \to 0}$ $[( \sin x \cos h +\cos x \sin h)^2 – \sin^2x]/h$  by the formula $\sin(x+h)$ $= \sin x \cos h +\cos x \sin h$.

$= \lim\limits_{h \to 0}$  $(\sin^2 x \cos^2 h +2\sin x \cos h\cos x \sin h$ $+\cos^2x\sin^2h – \sin^2x)$/h,  as we know that $(a+b)^2$ $=a^2+2ab+b^2$

$= \lim\limits_{h \to 0}$  $[\dfrac{\sin^2 x (\cos^2 h-1)}{h}$ $+ \dfrac{2\sin x \cos h\cos x \sin h}{h}$ $+\dfrac{\cos^2x\sin^2h}{h}]$

Let us now apply the Sum rule of limits.

$= \lim\limits_{h \to 0}$  $\dfrac{\sin^2 x (\cos^2 h-1)}{h}$ $+ \lim\limits_{h \to 0}$ $\dfrac{2\sin x \cos h\cos x \sin h}{h}$ $+ \lim\limits_{h \to 0}$ $\dfrac{\cos^2x\sin^2h}{h}$

$=\sin^2 x \lim\limits_{h \to 0}$  $\dfrac{(\cos h-1)(\cos h+1)}{h}$ $+2 \lim\limits_{h \to 0}$ $\left(\sin x \cos h\cos x\dfrac{\sin h}{h} \right)$ $+ \lim\limits_{h \to 0}$ $\left( \cos^2 x \sin h \dfrac{\sin h}{h} \right)$

Now, applying the limit of sinh/h is 1 when h tends to 0, the above is

$=\sin^2 x (\cos 0+1) \lim\limits_{h \to 0}$  $\dfrac{\cos h-1}{h}$ $+2\sin x \cos 0\cos x \cdot 1$ $+ \cos^2 x \sin 0 \cdot 1$

$=2 \sin^2 x \lim\limits_{h \to 0}$  $\dfrac{\cos h-1}{h}$ $+2\sin x\cos x$ $+ 0$

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

So the above quantity is equal to

$2 \sin^2 x \cdot 0$ $+2\sin x\cos x$ $+ 0$

$=0+2\sin x\cos x+ 0$

$=2\sin x \cos x$

So the derivative of sin square x is equal to 2sin x cos x.

Remark: Applying the trigonometric formula of sin 2x=2sin x cos x, we obtain that the derivative of sin^2x is sin 2x.