Question

evaluate (differentiate)
y' when x=2 and
y= 1/√(2x)​

Answer 1

Answer:

$\frac{-1}{8\sqrt{2} }$

Step-by-step explanation:

Given that,  $y = \frac{1}{\sqrt{2x} }$

           =>  $y = \frac{1}{2}(x)^{-\frac{1}{2} }$

Therefore,  $\frac{dy}{dx}= \frac{d}{dx}[\frac{1}{2}(x)^{-\frac{1}{2} }]$

                        $= \frac{1}{2} [-\frac{1}{2}*(x)^{-\frac{3}{2} } ]$

                   $\frac{dy}{dx}=\frac{-1}{4x\sqrt{x} }$

At x = 2 ,   $\frac{dy}{dx}=\frac{-1}{4(2)\sqrt{2} }$

                      $= \frac{-1}{8\sqrt{2} }$.