How do I use the chain rule to find the rate at which a bullet is rising? A: The equation of the path of a bullet fired into the air is $y=-20x(x-20)$ where x and y are displacements in metres horizontally and vertically from the origin. The bullet is moving horizontally at a constant rate 0.5m/s. Find the rate at which the bullet is rising when x=8. Answer = 40m/s So wrote the equation using the chain rule $\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}$ so $\frac{dy}{dt}=(-40x+40)\frac{1}{2}$ but when i sub in x=8 the answer is incorrect. Am I doing something wrong?

You are correct to use the chain rule. However, you seem to be writing out too few steps which is why it is difficult for you to identify where you made a mistake and to fix it.

Here's a general piece of advice -- don't rush and try to do everything in your head -- get used to writing every step out first. After writing every step out a million times you'll be more likely to do it in your head correctly.

Anyway, by the product rule $\frac{dy}{dx}=-20(x-20) + -20x(1)=-20x +400 -20x = -40x +400$. Because derivatives represent the rate of change of a quantity, we have that $\frac{dx}{dt}=0.5$.

Thus $\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}= (0.5)(-40x + 400) = -20x + 200$. At $x=8$ this equals $$-20(8) + 200 = -160 + 200 = 40$$