How do you find the derivative of the function #g(x) = sqrt(1 - 121 x^2)(arccos)(11 x)#? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 1 Answer Massimiliano Apr 15, 2015 In this way: #y'=1/(2sqrt(1-121x^2))*(-242x)*arccos(11x)+# #+sqrt(1-121x^2) * (-1/sqrt(1-121x^2) *11)=# #=-(121x)/sqrt(1-121x^2)arccos(11x)-11#. 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 6175 views around the world You can reuse this answer Creative Commons License