Question

find the derevitive with respect to x and give answer with expliannation​

Answer 1

Answer:

$\displaystyle\dfrac{d}{dx}\left(\dfrac{1}{\sin^2x-\cos^2x}\right)=\dfrac{4\cos(x)\sin(x)}{(\sin^2x-\cos^2x)^2}$

Step-by-step explanation:

Given,

$\displaystyle\dfrac{d}{dx}\left(\dfrac{1}{\sin^2x-\cos^2x}\right)$

We know,

$\dfrac{d}{dx}\left(\dfrac{1}{f(x)}}\right) = -\dfrac{\frac{d}{dx}f(x)}{(f(x))^2}$

Using this identity,

$\dfrac{d}{dx}\left(\dfrac{1}{\sin^2x-\cos^2x}}\right) = -\dfrac{\frac{d}{dx}\left(\sin^2x-cos^2x)}{(\sin^2x-\cos^2x)^2}$

$= -\dfrac{\frac{d}{dx}\left(\sin^2x)-\frac{d}{dx}(cos^2x)}{(\sin^2x-\cos^2x)^2}$

$=- \dfrac{2\sin(x)\cos(x) - 2\cos(x)(-\sin x)}{(\sin^2x-\cos^2x)^2}\\\\=\dfrac{4\cos(x)\sin(x)}{(\sin^2x-\cos^2x)^2}$