`xa n dy` are the sides of two squares such that `y=x-x^2` . Find the rate of the change of the area of the second square with respect to the first square.

Answer

To solve the problem, we need to find the rate of change of the area of the second square with respect to the area of the first square. Let's break this down step by step.

### Step 1: Define the areas of the squares
Let:
- \( A_1 \) be the area of the first square with side length \( x \).
- \( A_2 \) be the area of the second square with side length \( y \).

From the problem, we know:
\[ A_1 = x^2 \]
\[ A_2 = y^2 \]

### Step 2: Substitute the expression for \( y \)
We are given that:
\[ y = x - x^2 \]
Thus, we can express \( A_2 \) in terms of \( x \):
\[ A_2 = (x - x^2)^2 \]

### Step 3: Differentiate the areas with respect to \( x \)
Now, we differentiate both areas with respect to \( x \).

**Differentiating \( A_1 \)**:
\[ \frac{dA_1}{dx} = \frac{d}{dx}(x^2) = 2x \]

**Differentiating \( A_2 \)**:
Using the chain rule, we differentiate \( A_2 \):
\[ A_2 = (x - x^2)^2 \]
Let \( u = x - x^2 \). Then,
\[ A_2 = u^2 \]
Using the chain rule:
\[ \frac{dA_2}{dx} = 2u \cdot \frac{du}{dx} \]
Now, we need to find \( \frac{du}{dx} \):
\[ u = x - x^2 \implies \frac{du}{dx} = 1 - 2x \]
Substituting back:
\[ \frac{dA_2}{dx} = 2(x - x^2)(1 - 2x) \]

### Step 4: Find the rate of change of \( A_2 \) with respect to \( A_1 \)
We want to find \( \frac{dA_2}{dA_1} \):
Using the chain rule:
\[ \frac{dA_2}{dA_1} = \frac{\frac{dA_2}{dx}}{\frac{dA_1}{dx}} \]
Substituting the derivatives we found:
\[ \frac{dA_2}{dA_1} = \frac{2(x - x^2)(1 - 2x)}{2x} \]
Simplifying:
\[ \frac{dA_2}{dA_1} = \frac{(x - x^2)(1 - 2x)}{x} \]

### Step 5: Final expression
Thus, the rate of change of the area of the second square with respect to the area of the first square is:
\[ \frac{dA_2}{dA_1} = (1 - x)(1 - 2x) \]