Question

hy guys please tell dy/dx of
4/x^2​

Answer 1

Answer:

$\\\tt \dfrac{-8}{x^3}$

Explanation:

Assuming y =  $\\\tt \dfrac{4}{x^2}$

$\\\tt \dfrac{dy}{dx} = \dfrac{d( \dfrac{4}{x^2} )}{dx} \\\tt = \dfrac{d(4x^{-2})}{dx}$

Using rule:

$\\\tt \dfrac{dx^n}{dx} = nx^{n-1}$

We'll get:

$\\\tt \dfrac{d(4x^{-2})}{dx} = -2(4)(x^{-3}) = -8x^{-3}$