Question

If \( \sqrt{x-xy} + \sqrt{y-xy} = 1 \), then \( \frac{dy}{dx} = \)

• A. \( -\sqrt{\frac{y-y^2}{x-x^2}} \)
• B. \( -\sqrt{\frac{1-y^2}{1-x^2}} \)
• C. \( -\sqrt{\frac{1-y}{1-x}} \)
• D. \( -\sqrt{\frac{x-y}{x+y}} \)

Hint:

For expressions like \( \sqrt{x(1-y)} + \sqrt{y(1-x)} = C \), a useful substitution is \( x=\sin^2 A \) and \( y=\sin^2 B \). This transforms the expression to \( \sin A \cos B + \cos A \sin B = \sin(A+B) \). Here, \( \sin(A+B)=1 \implies A+B = \pi/2 \). So \( \arcsin\sqrt{x} + \arcsin\sqrt{y} = \pi/2 \). Differentiating this implicitly often simplifies finding \( dy/dx \). The derivative of \( \arcsin u \) is \( \frac{1}{\sqrt{1-u^2}} \frac{du}{dx} \).

Answer 1

The Correct Option is A

The given equation is \( \sqrt{x(1-y)} + \sqrt{y(1-x)} = 1 \).
Let \( x = \sin^2 A \) and \( y = \sin^2 B \).
(This substitution is common for expressions involving \( \sqrt{x(1-x)} \)).
Then \( 1-x = \cos^2 A \) and \( 1-y = \cos^2 B \).
Assume \( x, y \in (0,1) \) so that \( \sin A, \cos A, \sin B, \cos B \) are positive.
The equation becomes: \( \sqrt{\sin^2 A \cos^2 B} + \sqrt{\sin^2 B \cos^2 A} = 1 \) \( |\sin A \cos B| + |\sin B \cos A| = 1 \) Assuming positive values, \( \sin A \cos B + \cos A \sin B = 1 \).
This is \( \sin(A+B) = 1 \).
Since A and B are angles corresponding to x and y (likely acute if using this substitution for simplicity), we can take \( A+B = \frac{\pi}{2} \).
So, \( B = \frac{\pi}{2} - A \).
Then \( \sin B = \sin(\frac{\pi}{2}-A) = \cos A \).
And \( \cos B = \cos(\frac{\pi}{2}-A) = \sin A \).
We have \( \sqrt{y} = \sin B = \cos A = \sqrt{1-\sin^2 A} = \sqrt{1-x} \).
So, \( \sqrt{y} = \sqrt{1-x} \implies y = 1-x \).
If \( y = 1-x \), then \( \frac{dy}{dx} = -1 \).
Let's check if the options give -1.
Option (1): \( -\sqrt{\frac{y-y^2}{x-x^2}} = -\sqrt{\frac{y(1-y)}{x(1-x)}} \).
If \( y=1-x \), then \( 1-y = x \).
So, \( -\sqrt{\frac{(1-x)x}{x(1-x)}} = -\sqrt{1} = -1 \).
This matches.
The substitution method seems effective.
Let's try implicit differentiation directly on \( \sqrt{x-xy} + \sqrt{y-xy} = 1 \).
Differentiating w.
r.
t.
x: \( \frac{1}{2\sqrt{x-xy}} \frac{d}{dx}(x-xy) + \frac{1}{2\sqrt{y-xy}} \frac{d}{dx}(y-xy) = 0 \) \( \frac{1}{2\sqrt{x(1-y)}} (1 - (1\cdot y + x \frac{dy}{dx})) + \frac{1}{2\sqrt{y(1-x)}} (\frac{dy}{dx} - (1\cdot y + x \frac{dy}{dx})) = 0 \) \( \frac{1-y-x\frac{dy}{dx}}{2\sqrt{x(1-y)}} + \frac{\frac{dy}{dx}(1-x)-y}{2\sqrt{y(1-x)}} = 0 \) Multiply by 2: \( \frac{1-y}{\sqrt{x(1-y)}} - \frac{x}{\sqrt{x(1-y)}}\frac{dy}{dx} + \frac{1-x}{\sqrt{y(1-x)}}\frac{dy}{dx} - \frac{y}{\sqrt{y(1-x)}} = 0 \) \( \sqrt{\frac{1-y}{x}} - \sqrt{\frac{x}{1-y}}\frac{dy}{dx} + \sqrt{\frac{1-x}{y}}\frac{dy}{dx} - \sqrt{\frac{y}{1-x}} = 0 \) \( \frac{dy}{dx} \left( \sqrt{\frac{1-x}{y}} - \sqrt{\frac{x}{1-y}} \right) = \sqrt{\frac{y}{1-x}} - \sqrt{\frac{1-y}{x}} \) \( \frac{dy}{dx} \left( \frac{\sqrt{(1-x)(1-y)} - \sqrt{xy}}{\sqrt{y(1-y)}} \right) = \frac{\sqrt{xy} - \sqrt{(1-x)(1-y)}}{\sqrt{x(1-x)}} \) From \( \sin(A+B)=1 \), we had \( \sin A \cos B + \cos A \sin B = 1 \).
\( \sqrt{x}\sqrt{1-y} + \sqrt{1-x}\sqrt{y} = 1 \).
So \( \sqrt{x(1-y)} + \sqrt{y(1-x)} = 1 \).
The term \( \sqrt{(1-x)(1-y)} - \sqrt{xy} \) is \( \cos A \cos B - \sin A \sin B = \cos(A+B) \).
Since \(A+B=\pi/2\), \( \cos(A+B)=0 \).
So, \( \sqrt{(1-x)(1-y)} = \sqrt{xy} \).
Then the terms in brackets become 0.
This implies \( \frac{dy}{dx} \cdot 0 = 0 \), which is not helpful.
This happens because \(y=1-x\) means \(x+y=1\).
The relation derived from substitution is indeed simpler.
The substitution \( x=\sin^2 A, y=\sin^2 B \) simplifies the expression to \( \sin(A+B)=1 \).
If \( A+B = \pi/2 \), then \( \arcsin\sqrt{x} + \arcsin\sqrt{y} = \pi/2 \).
Differentiating w.
r.
t.
x: \( \frac{1}{\sqrt{1-x}} \cdot \frac{1}{2\sqrt{x}} + \frac{1}{\sqrt{1-y}} \cdot \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0 \) \( \frac{1}{2\sqrt{x(1-x)}} + \frac{1}{2\sqrt{y(1-y)}} \frac{dy}{dx} = 0 \) \[ \frac{dy}{dx} = -\frac{2\sqrt{y(1-y)}}{2\sqrt{x(1-x)}} = -\sqrt{\frac{y(1-y)}{x(1-x)}} = -\sqrt{\frac{y-y^2}{x-x^2}} \] This directly matches option (1).