How do you differentiate #y=sec^-1(4x)#? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 1 Answer trosk Apr 21, 2017 #1/(x sqrt(16x^2 - 1))# Explanation: #sec(y(x)) = 4x => d/dx sec(y(x)) = 4# But #d/dx sec(y(x)) = sec^2(y(x)) * sin(y(x)) * y'(x)# So #y'(x) = 4/(sec^2(y(x)) * sin(y(x)))# But note that: #sec(y(x)) = 4x# #sin(y(x)) = sqrt (1 - cos(y(x))^2) = sqrt (1-(1/(4x))^2)# Hence #y'(x) = 4/(sec^2(y(x)) * sin(y(x))) = 1/(x sqrt(16x^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 2398 views around the world You can reuse this answer Creative Commons License