How do I find f'(x) for f(x)=4x^3+x^2+5x+8?

This question require you to differentiate. The general rule for this format of differentiation is to multiply the number in front of the x by the power of the x and then reduce the power by one (in formula terms this is f'(x)=nax^n-1). It's that simple. So for 4x^3 the number in front of the x is 4 and 3 is the power of the x. So 43=12. Then the power of the x, the 3, reduces by one becoming a 2. Therefore, the result is 12x^2. You would then do the same thing for the remaining parts of the equation. x^2 becomes 2x: x^2 is the same as 1x^2 so we do the number in front multiplied by the power (12=2) and then reduce the power by one (2-1=1). Therefore we get 2x^1 = 2x. 5x becomes simply 5: 5x=5x^1, multiply front: 5*1=5, reduce power: 1-1=0. As x^0=1, 5x^0= 5. 8 disappears when differentiated: This is because 8=8x^0 so 8x0=0. So overall when differentiating this equation we get f'(x) = 12x^2 + 2x + 5.