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
$\frac{dy}{dx}=\frac{2}{x+\sqrt{x^2+25}}\left(1+\frac{1}{2\sqrt{x^2+25}}(2x)\right)$
$=\frac{2}{x+\sqrt{x^2+25}}\left(\frac{\sqrt{x^2+25}+x}{\sqrt{x^2+25}}\right)$
$=\frac{2}{\sqrt{x^2+25}}$