What is the derivative of #y=ln(x^4sin^2x)#? Calculus Basic Differentiation Rules Chain Rule 1 Answer Ratnaker Mehta Jul 14, 2016 #dy/dx=4/x+2cotx.# Explanation: We first simplify #y# using the Rules of Log. #y=ln(x^4*sin^2x)=lnx^4+lnsin^2x=4lnx+2lnsinx# #:. dy/dx=(4lnx)'+(2lnsinx)'# #=4*1/x+2*1/sinx*(sinx)'#.................[Chain Rule] #=4/x+2/sinx*cosx# #:. dy/dx=4/x+2cotx.# 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 2987 views around the world You can reuse this answer Creative Commons License