How do I find the derivative of #y=arctan(cos θ)#? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 1 Answer Michael Jun 1, 2015 #y'=-(sintheta)/(cos^(2)theta+1)# #y=arctan(costheta)# Applying the chain rule: #y'=(1)/(cos^(2)theta+1)xx(d[costheta])/("d"theta)# #=(-sintheta)/(cos^(2)theta+1)# #=-(sintheta)/(cos^(2)theta+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 3680 views around the world You can reuse this answer Creative Commons License