Let x and y be the sides of two squares such | vedclass

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(A) Let $A_1$ be the area of the first square and $A_2$ be the area of the second square.Given the side of the first square is $x$,its area is $A_1 =

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$1 - 3 x + 2 x^2$

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(A) Let $A_1$ be the area of the first square and $A_2$ be the area of the second square.Given the side of the first square is $x$,its area is $A_1 = x^2$.Given the side of the second square is $y = x - x^2$,its area is $A_2 = y^2 = (x - x^2)^2$.Expanding this,we get $A_2 = x^2 - 2 x^3 + x^4$.We need to find the rate of change of $A_2$ with respect to $A_1$,which is $\frac{d A_2}{d A_1}$.Using the chain rule,$\frac{d A_2}{d A_1} = \frac{d A_2 / d x}{d A_1 / d x}$.First,calculate the derivatives with respect to $x$:$\frac{d A_1}{d x} = \frac{d}{d x}(x^2) = 2 x$.$\frac{d A_2}{d x} = \frac{d}{d x}(x^2 - 2 x^3 + x^4) = 2 x - 6 x^2 + 4 x^3$.Now,substitute these into the chain rule formula:$\frac{d A_2}{d A_1} = \frac{2 x - 6 x^2 + 4 x^3}{2 x} = \frac{2 x(1 - 3 x + 2 x^2)}{2 x} = 1 - 3 x + 2 x^2$.

Let $x$ and $y$ be the sides of two squares such that $y = x - x^2$. The rate of change of the area of the second square with respect to the area of the first square is

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