<s>[INST] The shortest distance from a point to the graph of the function To compute the distance from the point (5,5) to the graph of xy=4. I choose an arbitrary point (u,v) on the graph of $xy=4$.
 I get $d(u,v)=\sqrt{(u-5)^2+(v-5)^2}$ again $(u,v)$ satisfies equation of hyperbola so that $uv=4$. Now what shall i do next?
 [/INST] You're after the minimum of $f(x)=(x-5)^2+\left(\frac4x-5\right)^2$, with $x\in\mathbb R$. Now,\begin{align}f'(x)&=-10-\frac{32}{x^3}+\frac{40}{x^2}+2x\\&=2\frac{x^4-5x^3+20x-16}{x^3}\\&=2\frac{(x^2-4)(x^2-5x+4)}{x^3}.\end{align}So, see what happens at the roots of $f'(x)$, which are $\pm2$, $1$ and $4$.
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