How do you take the derivative of #y=tan^-1 sqrt(3x)#? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 1 Answer Lucy Jul 14, 2018 #(dy)/(dx)=sqrt(3x)/(2x(1+3x))# Explanation: #y=tan^(-1)sqrt(3x)# Recall if #y=tan^(-1)x#, then #(dy)/(dx)=1/(1+x^2)# #(dy)/(dx)=1/(1+(sqrt(3x))^2)times (sqrt3/(2sqrtx))# #(dy)/(dx)=1/(1+3x)timessqrt(3x)/(2x)# #(dy)/(dx)=sqrt(3x)/(2x(1+3x))# 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 1834 views around the world You can reuse this answer Creative Commons License