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 are on a right track. You have found the derivative and now just plug $x=5$ in it and you have:
$$f'(x) = \frac{2x-1}{2\sqrt{x^2-x+5}}=\frac{10-1}{2\sqrt{25-5+2}} = \frac{9}{10}$$
This gives you the slope of the tangent at the point $(5,5)$ so you have $y=\frac 9{10}x + b$. But note that $(5,5)$ is a point that lies on both the tangent line and the function graph so you have:
$$5 = \frac 9{10}\cdot 5 + b \implies b = 5 - \frac 92 = \frac 12$$
So the final result is:$y = \frac 9{10}x + \frac 12$
