Calculus of a Single Variable, Chapter 2, 2.3, Section 2.3, Problem 13

You need to first find derivative of the function using the product rule:
f'(x)= (x^3 + 4x)'*(3x^2+2x-5)+(x^3 + 4x)*(3x^2+2x-5)'
f'(x)= (3x^2 + 4)*(3x^2+2x-5)+(x^3 + 4x)*(6x+2)
f'(x)= 9x^4 + 6x^3 - 15x^2 + 12x^2 + 8x - 20 + 6x^4 + 2x^3 + 24x^2 +8x
Combining like terms yields:
f'(x)=15x^4 + 8x^3 + 21x^2 + 16x - 20
Hence, evaluating the derivative of the function, yields f'(x)=15x^4 + 8x^3 + 21x^2 + 16x - 20.
You need to evaluate f'(c) at c = 0, hence, you need to replace 0 for x in equation of f'(x).
f'(0)=15*0^4 + 8*0^3 + 21*0^2 + 16*0 - 20
f'(0)=-20
Hence, evaluating the derivative of the function, yields f'(x)=15x^4 + 8x^3 + 21x^2 + 16x - 20 and evaluating f'(c) at c = 0, yields f'(0)= -20.