What is the derivative of this function #sec^-1(x^2-x)#? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 1 Answer Shwetank Mauria Aug 5, 2016 #d/(dx)sec^(-1)(x^2-x)=(2x-1)/(|x^2-x|sqrt((x^2-x)^2-1))# Explanation: As derivative of #sec^(-1)x=1/(|x|sqrt(x^2-1))#, using chain rule, #d/(dx)sec^(-1)(x^2-x)=1/(|x^2-x|sqrt((x^2-x)^2-1))xx(2x-1)# = #(2x-1)/(|x^2-x|sqrt((x^2-x)^2-1))# Answer link Related questions What is the derivative of #f(x)=sin^-1(x)# ? What is the derivative of #f(x)=cos^-1(x)# ? What is the derivative of #f(x)=tan^-1(x)# ? What is the derivative of #f(x)=sec^-1(x)# ? What is the derivative of #f(x)=csc^-1(x)# ? What is the derivative of #f(x)=cot^-1(x)# ? What is the derivative of #f(x)=(cos^-1(x))/x# ? What is the derivative of #f(x)=tan^-1(e^x)# ? What is the derivative of #f(x)=cos^-1(x^3)# ? What is the derivative of #f(x)=ln(sin^-1(x))# ? See all questions in Differentiating Inverse Trigonometric Functions Impact of this question 1229 views around the world You can reuse this answer Creative Commons License