Question

Find dy/dx if y=1/x³​

Answer 1

Given :

We have been provided : $\bf{y=\dfrac{1}{x^3}}$

To Find :

We have to find dy/dx$.$

SoluTion :

$:\implies\sf\:y=\dfrac{1}{x^3}$

$:\implies\sf\:\dfrac{dy}{dx}=x^{-3}\dfrac{1}{dx}$

$:\implies\sf\:\dfrac{dy}{dx}=-3x^{-3-1}$

$:\implies\sf\:\dfrac{dy}{dx}=-3x^{-4}$

$:\implies\boxed{\bf{\purple{\dfrac{dy}{dx}=-\dfrac{3}{x^4}}}}$

❂ Remember :

$\dag\bf\:y=x^n$

$\dag\bf\:\dfrac{dy}{dx}=nx^{n-1}$

Answer 2

Given ,

The function is y = 1/x³ = x^(-3)

Differentiating y wrt x , we get

$\tt \implies \frac{dy}{dx} = \frac{d {(x)}^{ - 3} }{dx}$

$\tt \implies\frac{dy}{dx} = - 3 {(x)}^{ - 3 - 1}$

$\tt \implies \frac{dy}{dx} = - 3 {(x)}^{ - 4}$

$\tt \implies \frac{dy}{dx} = \frac{ - 3}{ {(x)}^{4} }$

Remmember :

$\tt \implies \frac{d {(x)}^{n} }{dx} = n {(x)}^{n - 1}$

$\tt \implies \frac{d(u \pm v)}{dx} = \frac{d(u)}{dx} \pm\frac{d(v)}{dx}$

$\tt \implies \frac{d(u.v)}{dx} = v \frac{d(u)}{dx} + u \frac{d(v)}{dx}$

$\implies \tt \frac{d}{dx} (\frac{u}{v} ) = \frac{v \frac{d(u)}{dx} - u \frac{d(v)}{dx} }{ {(v)}^{2} }$