Question #f87ec Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 1 Answer Jim S Nov 16, 2017 #f'(x)=4(sin/cos)'# #-6(6x)# #= 4((cos^2x+sin^2x)/cos^2x)# #- 36x# = #=4/cos^2x-36x# # , sin^2x+cos^2x = 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 1393 views around the world You can reuse this answer Creative Commons License