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 23609 views around the world You can reuse this answer Creative Commons License