f'(x)= (x²+1)(x+4) + x(2x)(x+4)+x(x²+1)

or you can multiply it all out and do the simple derivative of a polynomial

F(x) = x(x^2 + 1) (X + 4)

When expanded, it will give you the following.

F(x) = x(x^3 + 4x^2 + x + 4)

Furthermore,

F(x) = x^4 + 4x^3 + x^2 + 4x)

This time you can now get its derivative, that is F'(x).

F'(x) = 4x^3 + 12x^2 + 2x + 4

HOPE THIS HELPS

X (X² + 1) (X + 4) into X⁴+ 4X³ + X² + 4X using FOIL:

X (X² + 1) = X³ + 1X

(X³ + 1X) (X + 4) = 1X⁴+ 4X³ + 1X² + 4X

Then derive the simplified equation using the rule F´(x) = nXⁿ⁻¹

So:

1X⁴ becomes 4X³

4X³ becomes 12X²

1X² becomes 2X

4X becomes 4 (in this case we can think of 4X as 4X¹ thus 1 * 4X ¹⁻¹⁼⁰ = 4)

And all together you get F´(x) = 4X³ + 12X² + 2X + 4