# Analysis of Rational Functions (Using Calculus)

• Mar 22nd 2009, 12:19 PM
mike_302
Analysis of Rational Functions (Using Calculus)
Hi,

So the question is to analyze this rational function... I am easily able to get the first derivative. The second derivative causes my problems though because I don't have the chain rule! My first and most obvious question would be: This isn't an assignment... And we learn the chain rule early this week. Should I just ignore this question until we get the chain rule?

If the answer is no, then here is the first derivative (it is done correctly, as it was an example)

f'(x)= (-5x^2+5)/(x^2+1)^2

I guess, based on theory, that the best form to get this in, would be factored top and bottom. The example goes through one method and I can follow it for quite a while, but then it does one tricky thing that, as far as I can tell, would either be an ambiguous case, or else I don't understand it, so I am asking if someone here could go through the steps how how they would find the second derivative now.

Mike
• Mar 22nd 2009, 12:31 PM
running-gag
Hi

You managed to find the derivative of $\frac{5x}{x^2+1}$ using the formula $\left(\frac uv \right) '= \frac{u'v-uv'}{v^2}$

The derivative is $\frac{-5x^2+5}{(x^2+1)^2}$

The second derivative is obtained exactly the same way

Set $u(x) = -5x^2+5$ and $v(x) = (x^2+1)^2$ and use the above formula
• Mar 22nd 2009, 12:57 PM
mike_302
I understand that part... Problem is, it gets really difficult to get the final answer of the second derivative in factored form.

That's where I need someone to show me how that works out. Especially so with this case
• Mar 22nd 2009, 01:05 PM
running-gag
Using the formula I can get as second derivative

$\frac{10x(x^2-3)}{(x^2+1)^3}$
• Mar 22nd 2009, 01:46 PM
mike_302
I understand that much... And I have the answer available. But my problem is still the STEPS to get there. How did you get that in factored form? The steps in the example keep it in factored form, and again, the question is, is this an ambiguous case? A rare instance where it works that I can keep it in factored form?
• Mar 22nd 2009, 02:07 PM
skeeter
$f'(x) = \frac{-5x^2 + 5}{(x^2 + 1)^2}$

$f'(x) = -5 \cdot \frac{x^2-1}{(x^2 + 1)^2}$

$f''(x) = -5 \cdot \frac{(x^2+1)^2 \cdot 2x - (x^2-1) \cdot 4x(x^2+1)}{(x^2+1)^4}$

factor out $2x(x^2+1)$ from both terms in the numerator ...

$f''(x) = -5 \cdot \frac{2x(x^2+1)[(x^2+1) - 2(x^2-1)]}{(x^2+1)^4}$

combine like terms in the last factor of the numerator ...

$f''(x) = -5 \cdot \frac{2x(x^2+1)[3 - x^2]}{(x^2+1)^4}$

cancel the common factors of $(x^2+1)$ and clean up ...

$f''(x) = \frac{10x(x^2-3)}{(x^2+1)^3}$