Question

If $x=\frac{\left(1-\sqrt{y}\right)}{\left(1+\sqrt{y}\right)}$, then $\frac{dy}{dx}$is equal to

• A. $\frac{4}{{\left(x+1\right)}^2}$
• B. $\frac{4\left(x-1\right)}{{\left(x+1\right)}^3}$
• C. $\frac{\left(x-1\right)}{{\left(x+1\right)}^3}$
• D. $\frac{4}{{\left(x+1\right)}^3}$

Answer 1

The correct option is B

$\frac{4\left(x-1\right)}{{\left(x+1\right)}^3}$

Explanation for the correct option.

Given, $x=\frac{\left(1-\sqrt{y}\right)}{\left(1+\sqrt{y}\right)}$.

Find the value of $\frac{dy}{dx}$.

Step 1: First we find the value of y.

$\begin{array}{rcl}x& =& \frac{\left(1-\sqrt{y}\right)}{\left(1+\sqrt{y}\right)}\\ x\left(1+\sqrt{y}\right)& =& \left(1-\sqrt{y}\right)\\ x+x\sqrt{y}& =& \left(1-\sqrt{y}\right)\\ x\sqrt{y}+\sqrt{y}& =& \left(1-x\right)\\ \sqrt{y}\left(1+x\right)& =& \left(1-x\right)\\ \sqrt{y}& =& \frac{\left(1-x\right)}{\left(1+x\right)}\\ y& =& \frac{{\left(1-x\right)}^2}{{\left(1+x\right)}^2}\end{array}$

Step 2: Now differentiate $y$ with respect to $x$.

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{d}{dx}\left[\frac{{\left(1-x\right)}^2}{{\left(1+x\right)}^2}\right]\\ & =& \left[\frac{{\left(1+x\right)}^2\times 2\times \left(1-x\right)\times \left(-1\right)-{\left(1-x\right)}^2\times \left(1+x\right)\times 2\times \left(1\right)}{{\left(1+x\right)}^4}\right]\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{[}\mathbf{\because }\frac{\mathbf{d}}{\mathbf{dx}}\mathbf{\left(}\frac{\mathbf{f}\left(x\right)}{\mathbf{g}\left(x\right)}\mathbf{\right)}\mathbf{=}\frac{\mathbf{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{f}\mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{-}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{g}\mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{{\mathbf{\left[}\mathbf{g}\left(x\right)\mathbf{\right]}}^{\mathbf{2}}}\mathbf{]}\\ & =& \left[\frac{2\left(1+x\right)\left(x-1\right)\left\{\left(1+x\right)-\left(x-1\right)\right\}}{{\left(1+x\right)}^4}\right]\\ & =& \left[\frac{2\left(1+x\right)\left(x-1\right)\left(1+1\right)}{{\left(1+x\right)}^4}\right]\\ & =& \frac{4\left(x-1\right)}{{\left(x+1\right)}^3}\end{array}$

Hence, the correct option is B.