What is the derivative of #y="arcsec"(x)#? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 1 Answer James May 19, 2018 #dy/dx=1/[x^2*sqrt(1-(1/x)^2)]# Explanation: show that #y=arcsecx=1/[arccosx]=arccos(1/x)# #d/dx[arccosu]=1/sqrt(1-u^2)*u'# #dy/dx=-1/[sqrt(1-(1/x)^2)]*[-1/x^2]# #dy/dx=1/[x^2*sqrt(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 23222 views around the world You can reuse this answer Creative Commons License