# [SOLVED] help with a derivative

• Jan 17th 2008, 08:46 AM
Discalculiated
[SOLVED] help with a derivative
ok, I'm in calc. concepts and I have two questions. Help with either would be greatly appreciated.

1)
Find an equation for the tangent line to f(x) = [-(6+3x^2)/x)] at (2, -9)
I know that the answer is 3x + 2y = -12 I just can't find it! I shouldn't have any problem finding the actual tan line, it's the derivatibe that I'm having trouble with.

I tried using quotient rule and this is what I got:

(-x)(-6x) - (-1)(6 + 3x^2)
------------------------- =
x^2

6x^2 + 6 + 3x^2
----------------------- =
x^2

9x^2 + 6
---------
x^2

which I don't think is right?

2) Find the derivative of
y = -6 - (1/x)
----------
X + 5

yikes! We don't have anything like this in the text, he just kind of snuck it in on the test and I'm trying to figure it out. The answer is:
6x^2 + 2x + 5
--------------
x^2(x+5)^2
• Jan 17th 2008, 11:30 AM
Peritus
$
f(x) = - \frac{{6 + 3x^2 }}{x}
$

if you choose to use the quotient rule:

$\begin{array}{l}
f(x) = \frac{{g(x)}}{{w(x)}} \\
f' (x) = \frac{{g' (x)w(x) - g(x)w' (x)}}{{w^2 (x)}} \\
\end{array}

$

in our case:

$\begin{array}{l}
g(x) = - 6 - 3x^2 - > g'(x) = - 6x \\
w(x) = x - > w'(x) = 1 \\
\end{array}$

thus:

$
f(x) = \frac{{ - 6x^2 + 6 + 3x^2 }}{{x^2 }} = - 3 + \frac{6}{{x^2 }}
$

-----------------------------------------------------------------------
First I'd suggest that you don't get fixated on finding special formulas for everything. The trick is to express the function in a convinient form and use the product rule.

$y(x) = \frac{{ - 6 - {\raise0.7ex\hbox{1} \!\mathord{\left/
{\vphantom {1 x}}\right.\kern-\nulldelimiterspace}
\!\lower0.7ex\hbox{x}}}}{{x + 5}} = - \left( {6 + \frac{1}{x}} \right) \cdot \left( {\frac{1}{{x + 5}}} \right)$

$
\begin{array}{l}
y'(x) = - \left( {\frac{1}{{x + 5}}} \right) \cdot \frac{d}{{dx}}\left( {6 + \frac{1}{x}} \right) - \left( {6 + \frac{1}{x}} \right)\frac{d}{{dx}}\left( {\frac{1}{{x + 5}}} \right) = \\
= \frac{1}{{x^2 }} \cdot \left( {\frac{1}{{x + 5}}} \right) + \left( {6 + \frac{1}{x}} \right) \cdot \left( {\frac{1}{{(x + 5)^2 }}} \right) \\
\end{array}
$

$
\begin{array}{l}
= \frac{1}{{(x + 5)x^2 }} + \frac{{6x + 1}}{{(x + 5)^2 x}} \\
{\rm common}\;{\rm denominator} \\
{\rm = }\frac{{{\rm x + 5 + 6x}^{\rm 2} + x}}{{x^2 (x + 5)^2 }} = \frac{{{\rm 6x}^{\rm 2} + 2x + 5}}{{x^2 (x + 5)^2 }} \\
\end{array}
$
• Jan 18th 2008, 09:34 PM
kalagota
after getting the correct answer in the first one, which is

$f'(x)=-3 + \frac{6}{x^2}$ = slope of the tangent line), all you have to do is to substitute x=2, and you'll get the slope of the line and just use the point-slope form to get the line..

--------------------------------
there are many ways to attack the problem..
just do the one you think you are good at..