How do you differentiate #f(x) =x(x+3)^3?# using the chain rule?

1 Answer
Oct 30, 2015

Answer:

I found: #f'(x)=(x+3)^2(4x+3)#

Explanation:

I would use the Chain and Product Rule to deal with the multiplication between the two functions #x# and #(x+3)^3#;
I would then use the Chain Rule (in red) to deal with #(x+3)^3# deriving first the cube leaving the argument as it is and multiplying the derivative of the argument inside the bracket:

#f'(x)=1*(x+3)^3+x*color(red)(3(x+3)^2*1)=#
#=(x+3)^2[(x+3)+3x]=#
#=(x+3)^2(4x+3)#