How do you differentiate #y=sec^-1(x^2)#? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 1 Answer sjc Jan 4, 2017 #(dy)/(dx)=2/(xsqrt(x^4-1))# Explanation: #y=sec^(-1)x^2# #=>x^2=secy# differentiate wrt #x# #2x=(dy)/(dx)secytany# #(dy)/(dx)=(2x)/(secytany)# now substitute back using: #secy=x^2# #sec^2y=1+tan^2y=>tany=sqrt(sec^2y-1# #tany=sqrt(x^4-1)# #(dy)/(dx)=(2x)/(x^2sqrt(x^4-1))=2/(xsqrt(x^4-1))# 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 7331 views around the world You can reuse this answer Creative Commons License