Derivatives?

Question

Let f(x) = x^2 (2x+1)^3 , then the derivative f(x) at x=1, f ' (1) = ?

Answer 1

Use the product rule and chain rule to find f'(x):

f(x) = x^2 (2x+1)^3
=> f'(x) = (d/dx x^2) (2x+1)^3 + (x^2) d/dx((2x+1)^3)
= 2x (2x+1)^3 + x^2 . 3(2x+1)^2. d/dx(2x + 1)
= 2x (2x+1)^3 + 3x^2 (2x+1)^2 . 2
= 2x (2x+1)^2 (2x+1 + 3x)
= 2x (2x+1)^2 (5x + 1).

So f'(1) = 2 (3^2) (6) = 108.

Answer 2

You use the product rule to determine the derivative- First (x^2) times derivative of second (6(2x+1)^2) plus the second (2x+1)^3 times the derivative of the first (2x).
= 6x^2(2x+1)^2 + 2x(2x +1)^3
Then you fill in 1 for x:
=6(2+1)^2 + 2(2 +1)^3 = 6(9)+2(27)= 108