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

1 Answer
Konstantinos Michailidis
Sep 12, 2015

It is #f'(x)=(2x)/((sqrt(x^2-1))*(x^2+1)^(3/2))#

Explanation:

We have that

#f(x)=sqrt((x^2-1)/(x^2+1))# hence

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

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

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

#f'(x)=(2x)/((sqrt(x^2-1))*(x^2+1)^(3/2))#

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