How do you differentiate #y = 6 cos(x^3 + 3)#? Calculus Basic Differentiation Rules Chain Rule 1 Answer marfre Mar 14, 2017 #y' = -18x^2 sin(x^3 + 3)# Explanation: Use #(cos u)' = -sin u (u')# Let #u = x^3 +3#, #u' = 3x^2# If #y = 6cos(x^3 + 3)# then #y' = 6 (-sin (x^3 + 3))(3x^2)# Simplify: #y' = -18x^2 sin(x^3 + 3)# 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 1999 views around the world You can reuse this answer Creative Commons License