Question

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

• A. 0
• B. -1
• C. 1
• D. 2

Answer 1

Answer: (A)

0

Answer 2

The correct answer is A:0
We have,
$f(x)=\sqrt{2 x}+\frac{4}{\sqrt{2 x}}=\sqrt{2 x}+4(2 x)^{-\frac{1}{2}}$
$\Rightarrow f^{'}(x)=\frac{1}{2 \sqrt{2 x}}-2+4\left[-\frac{1}{2}(2 x)^{-3 / 2}(2)\right]$
$=\frac{1}{\sqrt{2} \sqrt{x}}-\frac{4}{(2 x)^{3 / 2}}$
$=\frac{1}{\sqrt{2} \sqrt{x}}-\frac{4}{2 \sqrt{2} x^{3 / 2}}$
$=\frac{1}{\sqrt{2} \sqrt{x}}-\frac{\sqrt{2}}{x \sqrt{x}}$
Now $f(2)=\frac{1}{\sqrt{2}\times{\sqrt{2}}}-\frac{\sqrt{2}}{2\times\sqrt{2}}$
=0