Question

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

Hint:

Before jumping into implicit differentiation, which can be complicated, always check if the equation can be algebraically simplified to get an explicit function. This often leads to a much easier differentiation process.

Answer 1

Step 1: Understanding the Concept:
We are given an implicit relation between x and y and need to find its derivative, \( \frac{dy}{dx} \). While implicit differentiation is possible, it is often simpler to first algebraically manipulate the equation to express y explicitly as a function of x.
Step 2: Key Formula or Approach:
1. Algebraically simplify the given equation to solve for y in terms of x.
2. Differentiate the resulting explicit function \( y(x) \) using standard rules of differentiation (like the quotient rule).
Step 3: Detailed Explanation or Calculation:
The given equation is \( x\sqrt{1+y} + y\sqrt{1+x} = 0 \).
Rearrange the equation:
\[ x\sqrt{1+y} = -y\sqrt{1+x} \] Square both sides:
\[ x^2(1+y) = (-y)^2(1+x) \] \[ x^2 + x^2y = y^2 + y^2x \] Rearrange the terms to group x and y:
\[ x^2 - y^2 = y^2x - x^2y \] Factor both sides. The left side is a difference of squares, and we can factor -xy from the right side:
\[ (x-y)(x+y) = -xy(x-y) \] Assuming \( x \neq y \), we can divide both sides by \( (x-y) \):
\[ x+y = -xy \] Now, solve for y:
\[ y + xy = -x \] \[ y(1+x) = -x \] \[ y = -\frac{x}{1+x} \] Now we have an explicit function for y. We can differentiate it with respect to x using the quotient rule, \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \):
\[ \frac{dy}{dx} = -\left[ \frac{(1)(1+x) - (x)(1)}{(1+x)^2} \right] \] \[ \frac{dy}{dx} = -\left[ \frac{1+x-x}{(1+x)^2} \right] \] \[ \frac{dy}{dx} = -\frac{1}{(1+x)^2} \] Step 4: Final Answer:
Hence, it is proved that \( \frac{dy}{dx} = -\frac{1}{(1+x)^2} \).