How do you find the derivative of #y = tanh^-1 (1/x)#? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 1 Answer Steve M Nov 2, 2016 # dy/dx=1/(1-x^2) # Explanation: # y=tanh^-1(1/x) => tanhy=1/x # Differentiating implicitly gives: # (1-tanh^2y)dy/dx=-1/x^2 # # :. (1-(1/x)^2)dy/dx=-1/x^2 # # :. dy/dx=-1/((x^2)(1-(1/x)^2))# # :. dy/dx=-1/((x^2)(1-1/x^2))# # :. dy/dx=-1/(x^2-1)# # :. dy/dx=1/(1-x^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 5944 views around the world You can reuse this answer Creative Commons License