Find the closest point to the origin On the line,Find the closest point to the origin 
$y=\frac{2}{\sqrt{x}}$
What I did so far is : 
First point is : $(x,\frac{2}{\sqrt{x}})$ and point two is : (0,0)
$d=\sqrt{x^2+(\frac{2}{\sqrt{x}})^2}$
then - > $d' = 2x-\frac{4}{{x^2}}$
I found the X but what then? I do not think I did right.
Thanks.

You are almost there. First note that $y = \dfrac2{\sqrt{x}}$ is not a line. If $r$ is the distance from the origin, we have
$$r^2 = x^2 + \left(\dfrac2{\sqrt{x}} \right)^2 = x^2 + \dfrac4x$$
Now setting $\dfrac{dr^2}{dx} = 0$, we get that
$$2x - \dfrac4{x^2} = 0 \implies x^* =\sqrt[3]{2}$$
Hence, the value of $r^2$ at the point, which is at minimal distance from the origin is $$r^2 = 2^{2/3} + \dfrac4{2^{1/3}} = 2^{2/3} + 2^{5/3} = 3 \cdot 2^{2/3}$$
Hence, the minimum distance from the origin is
$$r = \sqrt{3} \cdot \sqrt[3]2$$
The point on the cuve is given by $$\left(x^*,\dfrac2{\sqrt{x^*}}\right) = \left(2^{1/3},\dfrac2{\sqrt[2]{2^{1/3}}}\right) = \left(2^{1/3}, 2^{5/6}\right)$$
