What is the derivative of #y= arctan (sqrt 3x^2 -1)#? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 1 Answer RedRobin9688 Jun 27, 2018 #y'=(2sqrt3x)/(1+(sqrt3x^2+1)^2)# Explanation: #color(blue)(d/dx)tan^-1u=((du)/dx)/(1+u^2)# Let #u=sqrt3x^2-1# , #(du)/dx=2sqrt3x# #y'=(2sqrt3x)/(1+(sqrt3x^2+1)^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 2115 views around the world You can reuse this answer Creative Commons License