How do you find the derivative of the function: #arccos(x^2)#? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 1 Answer Tom Apr 26, 2016 #theta =arccos(x^2)# #cos(theta) = x^2# #-theta'sin(theta) = 2x# #theta' = -(2x)/sin(theta)# don't forget that #sin(theta) = sqrt(1-cos^2(theta))# and that #cos(theta) = x^2# so we have #theta' = -(2x)/sqrt(1-x^4)# 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 1255 views around the world You can reuse this answer Creative Commons License