Question

If $y=log\left[\frac{x+\sqrt{x^{2}+25}}{\sqrt{x^{2}+25-x}}\right]$ then $\frac{dy}{dx}=......$

• A. $\frac{1}{\sqrt{x^{2}+25}}$
• B. $\frac{2}{\sqrt{x^{2}+25}}$
• C. $\frac{-1}{\sqrt{x^{2}+25}}$
• D. $\frac{-2}{\sqrt{x^{2}+25}}$

Answer 1

Answer: (B)

We have, $y=\log \left[\frac{x+\sqrt{x^{2}+25}}{\sqrt{x^{2}+25}-x}\right]$
$\Rightarrow y=\log \left[\frac{\left(x+\sqrt{x^{2}+25}\right)^{2}}{x^{2}+25-x^{2}}\right]$
$\Rightarrow y=\log \left[\frac{\left(x+\sqrt{x^{2}+25}\right)^{2}}{25}\right]$
$\Rightarrow y=2 \log \left(x+\sqrt{x^{2}+25}\right)-\log 25$
On differentiating both sides w.r.t. $x$, we get
$\frac{d y}{d x}=\frac{2}{x+\sqrt{x^{2}+25}}\left(1+\frac{1}{2 \sqrt{x^{2}+25}}(2 x)\right)$
$=\frac{2}{x+\sqrt{x^{2}+25}}\left(\frac{\sqrt{x^{2}+25}+x}{\sqrt{x^{2}+25}}\right)$
$=\frac{2}{\sqrt{x^{2}+25}}$