Is $\frac{d(A^{-2})}{dy} = \frac{1}{A^3} \frac{dA}{dy}$?

Question

Equation 9-4 in "Open-channel hydraulics" by Chow (1959) (a true classic, THE most important textbook for open-channel hydraulics) states (excluding an ever-present correction coefficient): $$\frac{d}{dy}(\frac{V^2}{2g})=\frac{Q^2}{2g}\frac{dA^{-2}}{dy}=-\frac{Q^2}{gA^3}\frac{dA}{dy}=-\frac{Q^2 T}{gA^3}$$ Where $V=Q/A$ (velocity equals discharge divided by area), $y$ is the depth, $g$ is gravitational acceleration, and $dA/dy=T$ (in a rectangular channel, $T$ is the width of the channel). What I don't get, is the third development, where $\frac{d(A^{-2})}{dy} = \frac{1}{A^3} \frac{dA}{dy}$. How can you put a variable just outside the derivative? Is that basic calculus?

Answer 1

It's chain rule, and the actual equality is $$\frac{d A^{-2}}{dy} = \frac{d A^{-2}}{dA} \frac{dA}{dy} = -2A^{-3} \frac{dA}{dy}.$$