Find the equation of a line tangent at a specific point I have to find an equation for the line tangent to the graph of
$\large\frac {\sqrt{x}}{6x+5}$
at the point $(4,f(4))$, and write it out in the form of $y=mx+b$
Using the quotient rule I get..
$(6x+5)\frac12 x^{-{\frac12}} - \large\frac{(6\sqrt{x})}{(6x+5)^2}$
I try plugging in $4$ for the slope and solving for "$b$" but it is not coming out correctly. 
I end up with..
$y=\frac{-4.75}{29^2}x+ \frac{2}{29}$
What am i doing wrong?

First you can simplify your derivative to: $$\dfrac{\mathrm d}{\mathrm dx}\left\{\dfrac{\sqrt{x}}{6x+5}\right\}=\dfrac{5-6x}{2\sqrt{x}(6x+5)^2},$$ which would make your calculations a little bit simpler. To find $b$ you just use the fact that $(4,f(4))$ lies in your tangent line, and you use the point-slope formula: $$y-y_0=m(x-x_0)\ \ \Rightarrow \ \ y-f(4)=f(4)(x-4) \ \ \Rightarrow \ \ \ldots $$
