Question

Y=(1/x^2+3)^2 diffrentiation

Answer 1

Answer:

$\displaystyle \sf \longrightarrow -\frac{12x^2+4}{x^5}$

Step-by-step explanation:

Given :

$\displaystyle \sf y= \left(\frac{1}{x^2} +3\right)^2$

We have to find y' :

Apply chain and power rule of differentiation :

Diff. w.r.t. x :

$\displaystyle \sf \longrightarrow y'=2. \left(\frac{1}{x^2} +3\right)^{2-1}. \left(\frac{1}{x^2} +3\right)'$

$\displaystyle \sf \longrightarrow y'=2. \left(\frac{1}{x^2} +3\right)^{}. \left(x^{-2} +3\right)'$

$\displaystyle \sf \longrightarrow y'=2. \left(\frac{1}{x^2} +3\right)^{}. \left(-2.x^{-2-1} +0\right)$

$\displaystyle \sf \longrightarrow y'=-2. \left(\frac{1}{x^2} +3\right)^{}. \left(2.x^{-3} \right)$

$\displaystyle \sf \longrightarrow y'=\frac{-4. \left(\dfrac{1}{x^2} +3\right)}{x^{-3} }$

$\displaystyle \sf \longrightarrow y'=-\frac{12x^2+4}{x^5}$

Hence we get required answer.