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
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.