If $x$ and $y$ are the sides of two squares such that $y=x-x^2$, find the rate of change of the area of the second square with respect to the first square.

 Given, $x$ and $y$ are sides of two squares, thus the area of two squares are $x^2$ and $y^2$.

The given curve is

$$\begin{gathered} y=x-x^{2 } \\ \Rightarrow \frac{dy}{dx}=1-2x \end{gathered}$$                                

Thus, the rate of change of area of first square $y^2$ with respect to $x^2$ is 

$\Rightarrow \frac{d\left(y^2\right)}{d\left(x^2\right)}=\frac{2y\frac{dy}{dx}}{2x}=\frac{y}{x}·\frac{dy}{dx}$

                 $=\frac{y}{x}\left(1-2x\right)$

                $$\begin{gathered} = \frac{\left(x-x^2\right)}{x}\left(1-2x\right) \\ = \left(1-x\right)\left(1-2x\right) \\ = 2x^2-3x+1 \end{gathered}$$

hence the rate of change of area of first square with respect to second square is $2x^2-3x+1$