The sides of two squares are $x$ and $y$ respectively, such that $y = x + x^{2}$ . The rate of change of area of second square with repsect to area of first square is

x^{2} + 3 x - 1

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2 x^{2} - 3 x + 1

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2 x^{2} + 3 x + 1

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1 + 2 x

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Solution

The correct option is C $2 x^{2} + 3 x + 1$
Let $A_{1} = x^{2}$ and $A_{2} = y^{2}$ be areas of two squares having sides as $x$ and $y$ respectively.
$\frac{d A_{2}}{d A_{1}} = \frac{\frac{d A_{2}}{d x}}{\frac{d A_{1}}{d x}}$
$\Rightarrow \frac{d A_{2}}{d A_{1}} = \frac{2 y \frac{d y}{d x}}{2 x}$
$= \frac{2 (x + x^{2}) (1 + 2 x)}{2 x}$
$= 2 x^{2} + 3 x + 1$

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