Question

If $f\left(x\right)=\sqrt{ax}+\frac{a^2}{\sqrt{ax}}$, then $f^{{}^{\prime }}\left(a\right)$ is equal to

• A. $-1$
• B. $1$
• C. $0$
• D. $a$

Answer 1

The correct option is C $0$

Given, $f\left(x\right)=\sqrt{ax}+\frac{a^2}{\sqrt{ax}}$
Then, $f^{{}^{\prime }}\left(x\right)=\frac{\sqrt{a}}{2\sqrt{x}}+\frac{a^2}{\sqrt{a}}(-\frac{1}{2}{x}^{-3/2})$
$\Rightarrow f^{{}^{\prime }}\left(x\right)=\frac{\sqrt{a}}{2\sqrt{x}}-\frac{a^2}{2\sqrt{a}}{x}^{-3/2}$
Therefore, $f^{{}^{\prime }}\left(a\right)=\frac{\sqrt{a}}{2\sqrt{a}}-\frac{a^2}{2\sqrt{a}{a}^{3/2}}$
$=\frac{1}{2}-\frac{a^2}{2a^2}=0$