What is the derivative of this function #y=29/4cot^-1((2x)/3)#? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 1 Answer James May 13, 2018 the answer #y'=-(58)/[12[1+((2x)/3)^2]]# Explanation: show below: #y=cot^-1(x)# #y'=[-1*x']/(1+(x)^2)# #y=29/4cot^-1((2x)/3)# #y'=29/4*-(2/3)/[1+((2x)/3)^2]# #y'=-(58/12)/[1+((2x)/3)^2]# #y'=-(58)/[12[1+((2x)/3)^2]]# 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 1326 views around the world You can reuse this answer Creative Commons License