Question

If $y=\log \left[\frac{\left(x+\sqrt{x^2+1}\right)}{\left(x+\sqrt{x^2-1}\right)}\right]$, then $\frac{d^{2}y}{d{x}^2}$ at $x=2$ is equal to

• A. $\begin{array}{rcl}& & 2\left(\frac{1}{3\sqrt{3}}-\frac{1}{5\sqrt{5}}\right)\end{array}$
• B. $\begin{array}{rcl}& & 2\left(\frac{1}{3\sqrt{3}}+\frac{1}{5\sqrt{5}}\right)\end{array}$
• C. $\begin{array}{rcl}& & \left(\frac{1}{3\sqrt{3}}-\frac{1}{5\sqrt{5}}\right)\end{array}$
• D. none of these

Answer 1

The correct option is A

$\begin{array}{rcl}& & 2\left(\frac{1}{3\sqrt{3}}-\frac{1}{5\sqrt{5}}\right)\end{array}$

Explanation for the correct option:

Step 1: Find $\frac{dy}{dx}$

Given that, $y=\log \left[\frac{\left(x+\sqrt{x^2+1}\right)}{\left(x+\sqrt{x^2-1}\right)}\right]$

Using the property of logarithmic, $\log \frac{a}{b}=\log a-\log b$ we have:

$y=\log \left(x+\sqrt{x^2+1}\right)-\log \left(x+\sqrt{x^2-1}\right)$

Now differentiate $y$ with respect to $x$ using chain rule.

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

Step 2: Find $\frac{d^{2}y}{d{x}^2}$

Now differentiate $\frac{dy}{dx}$ with respect to $x$we get:

$\begin{array}{rcl}\frac{d^{2}y}{d{x}^2}& =& -\frac{2x}{2{\left(x^2+1\right)}^{{\displaystyle \frac{3}{2}}}}+\frac{2x}{2{\left(x^2-1\right)}^{{\displaystyle \frac{3}{2}}}}\\ & =& -\frac{x}{{\left(x^2+1\right)}^{{\displaystyle \frac{3}{2}}}}+\frac{x}{{\left(x^2-1\right)}^{{\displaystyle \frac{3}{2}}}}\end{array}$

Put $x=2$ in $\frac{d^{2}y}{d{x}^2}$.

$\begin{array}{rcl}\frac{d^{2}y}{d{x}^2}& =& -\frac{2}{{\left(4+1\right)}^{{\displaystyle \frac{3}{2}}}}+\frac{2}{{\left(4-1\right)}^{{\displaystyle \frac{3}{2}}}}\\ & =& -\frac{2}{5^{{\displaystyle \frac{3}{2}}}}+\frac{2}{3^{{\displaystyle \frac{3}{2}}}}\\ & =& 2\left(-\frac{1}{5\sqrt{5}}+\frac{1}{3\sqrt{3}}\right)\end{array}$

Hence, option (A) is correct.