How do you find the derivative of # f(x)= 2/(1+x^2) #?

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Answer 1 · 2016-10-29T21:04:37

#y=f(x)#

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

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

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

#-2y^-2*(dy)/(dx)=2x#

#-2/(y^2)*(dy)/(dx)=2x#

Divide expressions on both sides of the equation by 2...

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

Multiply expressions on both sides of the equation by -1...

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

Now multiply expressions on both sides of the equation by y^2...

#(dy)/(dx)=-x*y^2#

Don't forget the real value of y...

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

Clean up the final result...

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

Which means that...

#f'(x)=-(4x)/((1+x^2)^2)#