How do you differentiate #f(x)=1/(1+x^2)^2#?

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Answer 1 · 2017-10-17T07:57:48

#dy/dx=(-4x)/(1+x^2)^3#

We have #y=1/(1+x^2)^2#

#y=(1+x^2)^-2#

Differentiate both sides using the chain rule.

#d/dxy=d/dx(1+x^2)^-2#

#dy/dx=(-2(1+x^2)^(-2-1))d/dx(1+x^2)#

#dy/dx=(-2(1+x^2)^-3)(2x^(2-1))#

#dy/dx=(-2(1+x^2)^-3)(2x)#

#dy/dx=(-2*2x)/(1+x^2)^3#

#dy/dx=(-4x)/(1+x^2)^3#