x and y are the sides of two squares such that y = x – x2. Find the rate of change of the area of the second square with respect to the area of the first square.

Given: two squares of sides x and y, such that y = x-x2


To find: the rate of change of the area of the second square with respect to the area of the first square


Explanation: Let the area of the first and the second square be A1 and A2 respectively.


Then Area of the first square is


A1 = x2


Differentiating this with respect to time, we get




And the area of the second square is


A2 = y2…….(ii)


But given, y = x-x2


Substituting this given value in equation (ii), we get


A2 = (x-x2)2 …….(iii)


Now differentiating equation (iii) with respect to t, we get



Now applying the power rule of differentiation, we get



Now applying the sum rule of differentiation, we get



Applying the derivative, we get





We need to find the rate of change of area of the second square with respect to the area of the first square, i.e.,



Substituting values from equation (i) and (iv), we get



By cancelling the like terms we get






Hence the rate of change of the area of the second square with respect to the area of the first square is 2x2-3x+1


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