Find f'(x) of this ? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 1 Answer Shwetank Mauria Mar 18, 2017 #f'(x)=cosxcotx-cscx-cotxcscx+12/x^5# Explanation: #f(x)=sinxcotx+cscx-3/x^4# Hence #f'(x)=d/(dx)(sinxcotx)+d/(dx)cscx-d/(dx)(3/x^4)# = #(cosxcotx+sinx xx(-csc^2x))-cotxcscx-3xx(-4)xx1/x^5# = #cosxcotx-cscx-cotxcscx+12/x^5# 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 1260 views around the world You can reuse this answer Creative Commons License