How do you find the derivative of the function #y = arccos(e^(5x))#? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 1 Answer Sonnhard May 29, 2018 #y'(x)=-1/sqrt(1-(e^(5x))^2)*exp(5x)*5# Explanation: Note that #(arc cos(x))'=-1/sqrt(1-x^2)# then we get by the chain rule #y'(x)=-1/sqrt(1-(e^(5x))^2)*5*e^(5x)# 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 6229 views around the world You can reuse this answer Creative Commons License