If x and y are sides of two squares such that | vedclass

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(C) Let $A_1$ be the area of the first square and $A_2$ be the area of the second square.Given that the side of the first square is $x$,so $A_1 = x^2$

Vedclass · Accepted Answer

$2(1 - 2x)(1 - 2x)$

Vedclass · Answer

(C) Let $A_1$ be the area of the first square and $A_2$ be the area of the second square.Given that the side of the first square is $x$,so $A_1 = x^2$.Given that the side of the second square is $y = x - x^2$,so $A_2 = y^2 = (x - x^2)^2$.We need to find the rate of change of $A_2$ with respect to $A_1$,which is $\frac{dA_2}{dA_1}$.Using the chain rule,$\frac{dA_2}{dA_1} = \frac{dA_2/dx}{dA_1/dx}$.First,calculate $\frac{dA_1}{dx} = \frac{d}{dx}(x^2) = 2x$.Next,calculate $\frac{dA_2}{dx} = \frac{d}{dx}((x - x^2)^2) = 2(x - x^2) \cdot \frac{d}{dx}(x - x^2) = 2(x - x^2)(1 - 2x)$.Now,substitute these into the chain rule formula:$\frac{dA_2}{dA_1} = \frac{2(x - x^2)(1 - 2x)}{2x} = \frac{2x(1 - x)(1 - 2x)}{2x} = (1 - x)(1 - 2x) = 1 - 2x - x + 2x^2 = 2x^2 - 3x + 1$.

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

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