How do you find the derivative of #sin(x cos x)#? Calculus Basic Differentiation Rules Chain Rule 1 Answer Salvatore I. Nov 26, 2016 By the chain rule #f'(x)=cos(xcosx)*(cosx-xsinx)# Explanation: #(df)/dx=(d(sin(g(x))))/dx=cos(g(x))*(dg(x))/dx# in our case #g(x)=x*cosx# so #g'(x)=cosx-xsinx# Answer link Related questions What is the Chain Rule for derivatives? How do you find the derivative of #y= 6cos(x^2)# ? How do you find the derivative of #y=6 cos(x^3+3)# ? How do you find the derivative of #y=e^(x^2)# ? How do you find the derivative of #y=ln(sin(x))# ? How do you find the derivative of #y=ln(e^x+3)# ? How do you find the derivative of #y=tan(5x)# ? How do you find the derivative of #y= (4x-x^2)^10# ? How do you find the derivative of #y= (x^2+3x+5)^(1/4)# ? How do you find the derivative of #y= ((1+x)/(1-x))^3# ? See all questions in Chain Rule Impact of this question 1373 views around the world You can reuse this answer Creative Commons License