How do you find the derivative of # y = cosx(cotx)#? Calculus Differentiating Trigonometric Functions Derivatives of y=sec(x), y=cot(x), y= csc(x) 1 Answer Euan S. Jul 20, 2016 #(dy)/(dx) = cosx(-csc^2x-1)# Explanation: We have to use the product rule. #d/(dx)(uv) = u(dv)/(dx) + (du)/(dx)v# #(dy)/(dx) = cosx(d/(dx)(cotx)) + d/(dx)(cosx)cotx# #= -cosxcsc^2x - sinxcotx# Remember that #cotx = 1/tanx = cosx/sinx# #therefore (dy)/(dx) = =-cosxcsc^2x - cosx# Answer link Related questions What is Derivatives of #y=sec(x)# ? What is the Derivative of #y=sec(x^2)#? What is the Derivative of #y=x sec(kx)#? What is the Derivative of #y=sec ^ 2(x)#? What is the derivative of #y=4 sec ^2(x)#? What is the derivative of #y=ln(sec(x)+tan(x))#? What is the derivative of #y=sec^2(x)#? What is the derivative of #y=sec^2(x) + tan^2(x)#? What is the derivative of #y=sec^3(x)#? What is the derivative of #y=sec(x) tan(x)#? See all questions in Derivatives of y=sec(x), y=cot(x), y= csc(x) Impact of this question 11438 views around the world You can reuse this answer Creative Commons License