chain rule derivative This is an derivatives Problem.

find the tangent to $y=\sqrt{x^2-x+5\;}\;$ at $x=5$.
$y = \boxed{\;?\;}$

What I did was first find y by plugin x into the equation. The answer is $y = 5$. Then I found the derivative of both $x^2-x+5$ and $\sqrt{x^2-x+5\;}\;$ the answers for those are $2x-1$ and $\tfrac 1 2 (x)^{-1/2}$. 
From here I know I have to find the answers by plugin in $5$ to $\tfrac 1 2 (x^2-x+5)^{-1/2} \cdot (2(x)-1)$  to find $m$.
After $y= mx +b$ to find $b$.
and help anyone Im on my last chance and really need the help.

You're almost there.   You have applied the chain rule to get the derivative.   You just have to substitute at the point: $x:=5$.
$\begin{align}
y &= \sqrt{x^2-x+5}
\\[2ex]
 {y}_x' & = {(x^2-x+5)}_x' \cdot {(u^{1/2})}_u'\mid_{u:=(x^2-x+5)}
\\[1ex] &= (2x-1) \cdot \tfrac 1 2 (x^2-x+5)^{-1/2}
\\[2ex]
 {y}_x'|_{x:=5} & = (2\cdot 5 - 1)\cdot \tfrac 1 2(5^5 -5+5)^{-1/2}
\\[1ex] & = \frac 9 {10} 
\end{align}$
Then since $y|_{x:=5} = \sqrt{5^2-5+5} = 5$ the tangent is $y_t -5 = \frac 9{10} (x_t-5)$ 
$$y_t = \frac 9 {10} x_t + \frac 1 2$$
