Question

find dy/dx of y=x^(-2/3)​

Answer 1

$\huge \boxed{ \boxed{ \red{\mathfrak{solution}}}}$

To find :

dy/dx

Given:

$y = {x}^{ - \frac{2}{3} }$

Differentiate w.r.t. x

$\implies\frac{dy}{dx} = \frac{d}{dx} ( {x}^{ - \frac{2}{3} } ) \\ \\ \implies \: \frac{dy}{dx} = - \frac{2}{3} {x}^{ - \frac{2}{3} - 1 } \\ \\ \implies \: \frac{dy}{dx} = - \frac{2}{3} {x}^{ \frac{ - 2 - 3}{3} } \\ \\ \implies \: \frac{dy}{dx} = - \frac{2}{3} {x}^{ - \frac{5}{3} }$

Formula used:

$\huge{ \boxed{ \green{\bold{ \frac{d}{dx} {x}^{n} = n {x}^{n - 1} }}}}$

Answer 2

$\huge\bold{Question}$

find dy/dx of y=x^(-2/3) ?

$\huge\bold{Answer}$

According to the question we have to find “ dy/dx ”

Given:

$\begin{lgathered}y = {x}^{ - \frac{2}{3} } \\ \\\end{lgathered}$

By using this Formula , we find the answer :-

$\huge\tt{ \frac{d}{dx} {x}^{n} = n {x}^{n - 1} }$

Now , we differentiate with respect to “ x ”

$\begin{lgathered}\tt\frac{dy}{dx} = \frac{d}{dx} ( {x}^{ - \frac{2}{3} } ) \\ \\\tt \frac{dy}{dx} = - \frac{2}{3} {x}^{ - \frac{2}{3} - 1 } \\ \\ \tt \frac{dy}{dx} = - \frac{2}{3} {x}^{ \frac{ - 2 - 3}{3} } \\ \\ \tt\frac{dy}{dx} = - \frac{2}{3} {x}^{ - \frac{5}{3} }\end{lgathered}$