x and y are the sides of two squares such that

Let’s consider the area of the first square A1 = x2

And, area of the second square be A2 = y2

Now, A1 = x2 and A2 = y2 = (x – x2)2

Differentiating both A1 and A2 w.r.t. t, we get

$\frac{d \mathrm{~A}_{1}}{d t}=2 x \cdot \frac{d x}{d t}$ and $\frac{d \mathrm{~A}_{2}}{d t}=2\left(x-x^{2}\right)(1-2 x) \cdot \frac{d x}{d t}$

Thus, $\frac{d \mathrm{~A}_{2}}{d \mathrm{~A}_{1}}=\frac{\frac{d \mathrm{~A}_{2}}{d t}}{\frac{d \mathrm{~A}_{1}}{d t}}=\frac{2\left(x-x^{2}\right)(1-2 x) \cdot \frac{d x}{d t}}{2 x \cdot \frac{d x}{d t}}$

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

$=1-2 x-x+2 x^{2}=2 x^{2}-3 x+1$

Therefore, the rate of change of area of the second square with respect to first is 2x2 – 3x + 1.