Question

If $f\left(x\right)=\frac{a+\sqrt{a^2-x^2}+x}{\sqrt{a^2-x^2}+a-x}$ where $a>0$ and $x<a$, then $f^{{}^{\prime }}\left(0\right)$ has the value equal to

• A. $\sqrt{a}$
• B. $a$
• C. $\frac{1}{\sqrt{a}}$
• D. $\frac{1}{a}$

Answer 1

The correct option is D $\frac{1}{a}$

We have
$f\left(x\right)=\frac{a+\sqrt{a^2-x^2}+x}{\sqrt{a^2-x^2}+a-x}$
$=\frac{(a+x)+\sqrt{a-x}\sqrt{a+x}}{(a-x)+\sqrt{a-x}\sqrt{a+x}}$
$=\frac{\sqrt{a+x}(\sqrt{a+x}+\sqrt{a-x})}{\sqrt{a-x}(\sqrt{a-x}+\sqrt{a+x})}$
$={\left(\frac{a+x}{a-x}\right)}^{1/2}$
Differentiating w.r. to $x$
$\frac{dy}{dx}=\frac{1}{2}\sqrt{\frac{a-x}{a+x}}\left[\frac{(a-x)+(a+x)}{{(a-x)}^2}\right]$
$\frac{dy}{dx}=\frac{1}{2}\frac{\sqrt{a-x}}{\sqrt{a+x}}\times \frac{2a}{{(a-x)}^2}$
$\frac{dy}{dx}=\frac{1}{a}$