Question

If $y=\frac{x}{2}\sqrt{a^2+x^2}+\frac{a^2}{2}\log [x+\sqrt{a^2+x^2}]$, then $\frac{dy}{dx}=?$

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

Answer 1

The correct option is A

$\sqrt{x^2+a^2}$

Explanation for the correct option:

Finding the value of $\frac{dy}{dx}$:

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

Differentiating both sides with respect to $x$ we get

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

Hence, option A is correct.