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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?

jrounsav
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1 Answers1

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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: $$y-y_0=m(x-x_0)\ \ \Rightarrow \ \ y-f(4)=f(4)(x-4) \ \ \Rightarrow \ \ \ldots $$

Hakim
  • 10,213
  • My bad, I made an edit to my post. The denominator in the slope of my answer was supposed to be to the 2nd power, which is a reduced form of the same slope you offered. The answer still does not work – jrounsav Jul 15 '14 at 03:52
  • @Cluckles Fixed, look at my edit. – Hakim Jul 15 '14 at 03:54