How do you find the derivative of #sqrt((x^2-1) / (x^2+1))#? Calculus Basic Differentiation Rules Chain Rule 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))# Answer link Related questions What is the Chain Rule for derivatives? How do you find the derivative of #y= 6cos(x^2)# ? How do you find the derivative of #y=6 cos(x^3+3)# ? How do you find the derivative of #y=e^(x^2)# ? How do you find the derivative of #y=ln(sin(x))# ? How do you find the derivative of #y=ln(e^x+3)# ? How do you find the derivative of #y=tan(5x)# ? How do you find the derivative of #y= (4x-x^2)^10# ? How do you find the derivative of #y= (x^2+3x+5)^(1/4)# ? How do you find the derivative of #y= ((1+x)/(1-x))^3# ? See all questions in Chain Rule Impact of this question 1571 views around the world You can reuse this answer Creative Commons License