Q105 / 160 · TS EAMCET 2009 · Mathematics

$x=\frac{1-\sqrt{y}}{1+\sqrt{y}} \Rightarrow \frac{d y}{d x}$ is equal to

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

Solution

Given, $\frac{x}{1}=\frac{1-\sqrt{y}}{1+\sqrt{y}}$ Applying componendo and dividendo, we get $\begin{array}{rlrl} & & \frac{1+x}{1-x}=\frac{(1+\sqrt{y})+(1-\sqrt{y})}{(1+\sqrt{y})-(1-\sqrt{y})} \\ \Rightarrow & \frac{1+x}{1-x}=\frac{2}{2 \sqrt{y}} \\ \Rightarrow & & y=\left(\frac{1-x}{1+x}\right)^2 \end{array}$ On differentiating w.r.t. $x$, we get $\begin{aligned} \frac{d y}{d x} & =\frac{-2(1+x)^2(1-x)-(1-x)^2 \cdot 2(1+x)}{(1+x)^4} \\ & =\frac{(1-x)(1+x)(-2-2 x-2+2 x)}{(1+x)^4} \\ & =\frac{4(x-1)}{(x+1)^3} \end{aligned}$

Hence, the correct answer is option: 'B'.

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