<s>[INST] 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.
 [/INST] 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 $y_t -5 = \frac 9{10} (x_t-5)$ 
$$y_t = \frac 9 {10} x_t + \frac 1 2$$
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