How do you differentiate #y = cos^3(3x+1)#? Calculus Differentiating Trigonometric Functions Derivative Rules for y=cos(x) and y=tan(x) 1 Answer GiĆ³ Apr 7, 2015 I would use the Chain Rule deriving the #()^3# first and then the #cos# and then the argument as: #y'=3cos^2(3x+1)*[-sin(3x+1)]*3=-9cos^2(3x+1)sin(3x+1)# Answer link Related questions What is the derivative of #y=cos(x)# ? What is the derivative of #y=tan(x)# ? How do you find the 108th derivative of #y=cos(x)# ? How do you find the derivative of #y=cos(x)# from first principle? How do you find the derivative of #y=cos(x^2)# ? How do you find the derivative of #y=e^x cos(x)# ? How do you find the derivative of #y=x^cos(x)#? How do you find the second derivative of #y=cos(x^2)# ? How do you find the 50th derivative of #y=cos(x)# ? How do you find the derivative of #y=cos(x^2)# ? See all questions in Derivative Rules for y=cos(x) and y=tan(x) Impact of this question 1907 views around the world You can reuse this answer Creative Commons License