Question

If \( x\sqrt{1+y} + y\sqrt{1+x} = 0\), \(-1 < x < 1\), then prove that \(\frac{dy}{dx} = -\frac{1}{(1+x)^2} \).

Hint:

For implicit functions involving square roots, squaring both sides can simplify the expression significantly. However, be mindful that squaring can introduce extraneous solutions, so it's good practice to check the context of the problem, as we did by discarding the \(y=x\) case.

Answer 1

Step 1: Understanding the Concept:
The problem asks to find the derivative of \(y\) with respect to \(x\) from an implicit equation. The strategy is to first simplify the equation algebraically to express \(y\) explicitly as a function of \(x\), and then differentiate.
Step 2: Key Formula or Approach:
The main approach involves algebraic manipulation to isolate \(y\), followed by differentiation using the quotient rule.
Quotient Rule: If \(y = \frac{u(x)}{v(x)}\), then \(\frac{dy}{dx} = \frac{u'v - uv'}{v^2}\).
Step 3: Detailed Explanation or Calculation:
Given the equation: \[ x\sqrt{1+y} + y\sqrt{1+x} = 0 \] Rearrange the terms: \[ x\sqrt{1+y} = -y\sqrt{1+x} \] Square both sides to eliminate the square roots: \[ (x\sqrt{1+y})^2 = (-y\sqrt{1+x})^2 \] \[ x^2(1+y) = y^2(1+x) \] Expand the terms: \[ x^2 + x^2y = y^2 + y^2x \] Rearrange the equation to group terms: \[ x^2 - y^2 + x^2y - y^2x = 0 \] Factorize the terms. The first two terms form a difference of squares, and we can factor \(xy\) from the last two terms: \[ (x-y)(x+y) + xy(x-y) = 0 \] Factor out the common term \((x-y)\): \[ (x-y)(x+y+xy) = 0 \] This implies either \(x-y = 0\) or \(x+y+xy = 0\).
If \(x-y=0\), then \(y=x\). Substituting this into the original equation gives \(x\sqrt{1+x} + x\sqrt{1+x} = 2x\sqrt{1+x} = 0\), which is only true for \(x=0\) or \(x=-1\). Since we are looking for a general relation for \(-1<x<1\), we consider the other case.
So, we use the relation: \[ x+y+xy = 0 \] Now, we express \(y\) as a function of \(x\): \[ y + xy = -x \] \[ y(1+x) = -x \] \[ y = -\frac{x}{1+x} \] Now, we differentiate \(y\) with respect to \(x\) using the quotient rule: \[ \frac{dy}{dx} = -\left[ \frac{\frac{d}{dx}(x) \cdot (1+x) - x \cdot \frac{d}{dx}(1+x)}{(1+x)^2} \right] \] \[ \frac{dy}{dx} = -\left[ \frac{1 \cdot (1+x) - x \cdot 1}{(1+x)^2} \right] \] \[ \frac{dy}{dx} = -\left[ \frac{1+x-x}{(1+x)^2} \right] \] \[ \frac{dy}{dx} = -\frac{1}{(1+x)^2} \] Step 4: Final Answer:
We have successfully shown that if \(x\sqrt{1+y} + y\sqrt{1+x} = 0\), then \(\frac{dy}{dx} = -\frac{1}{(1+x)^2}\).