What is the derivative of this function #y=xsin^-1x+sqrt(1-x^2)#? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 1 Answer Shwetank Mauria Aug 5, 2016 #(dy)/(dx)=sin^(-1)x# Explanation: As #y=xsin^(-1)x+sqrt(1-x^2)#, is sum of two functions, we can use product rule for first and chain rule for second. Hence, #(dy)/(dx)=x xx1/sqrt(1-x^2)+1xxsin^(-1)x+1/(2sqrt(1-x^2))xx(-2x)# = #x/sqrt(1-x^2)+sin^(-1)x-x/sqrt(1-x^2)# = #sin^(-1)x# 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 28009 views around the world You can reuse this answer Creative Commons License