Question

If $f(x)=\sqrt{2x}+\frac{4}{\sqrt{2x}}$ , then $f^{\prime }(2)$ is equal to

• A. 0
• B. -1
• C. 1
• D. 2
• E. -2

Answer 1

Answer: (A)

We have,
$f(x)=\sqrt{2x}+\frac{4}{\sqrt{2x}}=\sqrt{2x}+4(2x{)}^{-\frac{1}{2}}$
$\Rightarrow f^{{}^{\prime }}(x)=\frac{1}{2\sqrt{2x}}-2+4\left[-\frac{1}{2}(2x{)}^{-3/2}(2)\right]$
$=\frac{1}{\sqrt{2}\sqrt{x}}-\frac{4}{(2x{)}^{3/2}}$
$=\frac{1}{\sqrt{2}\sqrt{x}}-\frac{4}{2\sqrt{2}{x}^{3/2}}$
$=\frac{1}{\sqrt{2}\sqrt{x}}-\frac{\sqrt{2}}{x\sqrt{x}}$
$\therefore f^{{}^{\prime }}(2)=\frac{1}{\sqrt{2}\sqrt{2}}-\frac{\sqrt{2}}{2\sqrt{2}}$
$=\frac{1}{2}-\frac{1}{2}$
$=0$