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.
 A: it may be easier to work with $$y^2 = x^2 - x + 5 \tag 1$$ and you want to the tangent at $x = 5, y = 5.$  differentiating $(1),$  we get $$2y\, dy = 2x \, dx - 1 \,dx \to 10 dy = 10dx - dx= 9dx$$  and the tangent line is given by $$10(y-5) = 9(x-5) $$ this can be rearranged to look in the familiar form $$y = \frac9{10}x+\frac12. $$ 
A: 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:$y = \frac 9{10}x + \frac 12$
A: 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 $y_t -5 = \frac 9{10} (x_t-5)$ 
$$y_t = \frac 9 {10} x_t + \frac 1 2$$
A: You've got the value of $m$. Then by virtue of the fact that the tangent $y=mx+b$ must contain the point $(x_0,y_0)$ where $x_0=5,y_0=\sqrt{x_0^2-x_0+5}$, you can find $b$.
