Question:
\( x = \frac{1 - \sqrt{y}}{1 + \sqrt{y}} \implies \frac{dy}{dx} \) is equal to
\( x = \frac{1 - \sqrt{y}}{1 + \sqrt{y}} \implies \frac{dy}{dx} \) is equal to
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Use Componendo and Dividendo to isolate variables in rational expressions.
Updated On: Apr 10, 2026
- $\frac{4}{(x+1)^2}$
- $\frac{4(x-1)}{(1+x)^3}$
- $\frac{x-1}{(1+x)^3}$
- $\frac{4}{(x+1)^3}$
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The Correct Option is B
Solution and Explanation
Step 1: Express $y$ in terms of $x$
$x = \frac{1 - \sqrt{y}}{1 + \sqrt{y}}$. By Componendo and Dividendo: $\frac{1+x}{1-x} = \frac{2}{2\sqrt{y}} = \frac{1}{\sqrt{y}}$. $\sqrt{y} = \frac{1-x}{1+x} \implies y = \left(\frac{1-x}{1+x}\right)^2$.
Step 2: Differentiate
$\frac{dy}{dx} = 2\left(\frac{1-x}{1+x}\right) \cdot \frac{d}{dx}\left(\frac{1-x}{1+x}\right)$. $\frac{dy}{dx} = 2\left(\frac{1-x}{1+x}\right) \cdot \left[\frac{(1+x)(-1) - (1-x)(1)}{(1+x)^2}\right]$. $\frac{dy}{dx} = \frac{2(1-x)}{(1+x)} \cdot \frac{-2}{(1+x)^2} = \frac{4(x-1)}{(1+x)^3}$.
Final Answer: (b)
$x = \frac{1 - \sqrt{y}}{1 + \sqrt{y}}$. By Componendo and Dividendo: $\frac{1+x}{1-x} = \frac{2}{2\sqrt{y}} = \frac{1}{\sqrt{y}}$. $\sqrt{y} = \frac{1-x}{1+x} \implies y = \left(\frac{1-x}{1+x}\right)^2$.
Step 2: Differentiate
$\frac{dy}{dx} = 2\left(\frac{1-x}{1+x}\right) \cdot \frac{d}{dx}\left(\frac{1-x}{1+x}\right)$. $\frac{dy}{dx} = 2\left(\frac{1-x}{1+x}\right) \cdot \left[\frac{(1+x)(-1) - (1-x)(1)}{(1+x)^2}\right]$. $\frac{dy}{dx} = \frac{2(1-x)}{(1+x)} \cdot \frac{-2}{(1+x)^2} = \frac{4(x-1)}{(1+x)^3}$.
Final Answer: (b)
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