Find the smallest value of the following expression $\sqrt{(x-9)^2 +4} + \sqrt{x^2+y^2}+ \sqrt{(y-3)^2 +9}$ 
Find the smallest value of the following expression:
  $$\sqrt{(x-9)^2 +4} + \sqrt{x^2+y^2} + \sqrt{(y-3)^2 +9}$$

I tried to derive the expression with respect to $x$ and $y$ and then equal the derivative to zero and find the critical points, but I could not do that. It is complicated. Is there a way?

Proceeding as your first idea by derivative, we have that for
$$f(x,y)=\sqrt{(x-9)^2 +4} + \sqrt{x^2+y^2} + \sqrt{(y-3)^2 +9}$$
in order to minimize we can guess that $0<x<9$ and $0<y<3$ and then
$$f_x(x,y)=\frac{x}{\sqrt{x^2 +y^2}}+ \frac{x-9}{\sqrt{(x-9)^2+4}}=0 \implies y^2=\frac{4x^2}{(9-x)^2}\implies y=\frac{2x}{9-x}$$
$$f_y(x,y)=\frac{y}{\sqrt{x^2 +y^2}}+ \frac{y-3}{\sqrt{(y-3)^2+9}}=0\implies x^2=\frac{9y^2}{(3-y)^2}\implies x=\frac{3y}{3-y}$$
from which we obtain
$$(9-x)(3-y)=6$$
and then
$$y=\frac{2x}{(9-x)}\implies 3-\frac6{9-x}=\frac{2x}{(9-x)}$$
$$\implies {5x-21}{x-9}=0 \implies x=\frac{21}5$$
and 
$$y=\frac{2x}{(9-x)}=\frac{\frac{42}5}{\frac{24}5}=\frac 7 4$$
therefore the solution is $(x,y)=\left(\frac{21}5,\frac 7 4\right)$.
