How do you find the derivative of the function: #y = arccosx + x sqrt(1-x^2)#? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 1 Answer Leland Adriano Alejandro Jan 10, 2016 #dy/dx=(-2x^2)/sqrt(1-x^2# Explanation: #y=arccos x +xsqrt(1-x^2)# #dy/dx=-1/sqrt(1-x^2)+1*sqrt(1-x^2)+x*1/(2sqrt(1-x^2))*(-2x)# after simplification #dy/dx=(-2x^2)/sqrt(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 3128 views around the world You can reuse this answer Creative Commons License