Question

h tends to 0cos ^2(x+h)-cos^2x/h​

Answer 1

We have to simplify,

$\displaystyle\lim_{h\to0}\dfrac{\cos^2(x+h)-\cos^2x}{h}$

Actually this is the derivative of cos²x with respect to x.

Let's begin!

$\begin{aligned}&\lim_{h\to0}\dfrac{\cos^2(x+h)-\cos^2x}{h}\\\\\implies\ \ &\lim_{h\to0}\dfrac{(\cos(x+h)+\cos x)(\cos(x+h)-\cos x)}{h}\\\\\implies\ \ &\lim_{h\to0}\dfrac{2\cos\left(\dfrac{x+h+x}{2}\right)\cos\left(\dfrac{x+h-x}{2}\right)\cdot-2\sin\left(\dfrac{x+h+x}{2}\right)\sin\left(\dfrac{x+h-x}{2}\right)}{h}\\\\\implies\ \ &-\lim_{h\to0}\dfrac{2\sin\left(x+\dfrac{h}{2}\right)\cos\left(x+\dfrac{h}{2}\right)\cdot 2\sin\left(\dfrac{h}{2}\right)\cos\left(\dfrac{h}{2}\right)}{h}\end{aligned}$

$\begin{aligned}\implies\ \ &-\lim_{h\to0}\dfrac{\sin(2x+h)\cdot\sin h}{h}\\\\\implies\ \ &-\left[\lim_{h\to0}\ \sin(2x+h)\ \cdot\ \lim_{h\to0}\dfrac{\sin h}{h}\right]\\\\\implies\ \ &-\lim_{h\to0}\ \sin(2x)\cos h+\cos(2x)\sin h\quad\quad\left[\because\ \lim_{h\to0}\dfrac{\sin h}{h}=1\right]\\\\\implies\ \ &-\left[\sin(2x)\lim_{h\to0}\cos h+\cos(2x)\lim_{h\to0}\sin h\right]\\\\\implies\ \ &\boxed{-\sin(2x)}\end{aligned}$

Or we can simplify it as derivative of cos²x by applying product rule of differentiation.

$\displaystyle\lim_{h\to0}\dfrac{\cos^2(x+h)-\cos^2x}{h}\ =\ \frac{d}{dx}(\cos^2x)\\\\\\=\ \frac{d}{dx}(\cos x\cdot\cos x)\ =-\sin x\cos x-\cos x\sin x\\\\\\=\ -2\sin x\cos x\ =\ \boxed{-\sin(2x)}$