# Finding Derivative by Quotient Rule

• Jun 24th 2008, 03:15 PM
snakeman11689
Finding Derivative by Quotient Rule
$f(x)= (4x+x^2)/(2+x)^2$

I have to find the second prime derivative.
I got the first one, which resulted in the function in the above, but i have trouble doing the second derivative.

in the start i got....

$f(x)= (4+2x)(2+x)^2 - (4x+x^2)2(2+x)/ (2+x)^4$

From then on, i don't really know what to do. Can someone lecture me.
• Jun 24th 2008, 03:24 PM
Krizalid
There's no need to apply the quotient rule here, first let's make this thing up:

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

Now differerentiate.
• Jun 24th 2008, 03:28 PM
snakeman11689
Quote:

Originally Posted by Krizalid
There's no need to apply the quotient rule here, first let's make this thing up:

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

Now differerentiate.

What if i had to use the quotient rule...
• Jun 24th 2008, 04:06 PM
Reckoner
Quote:

Originally Posted by snakeman11689
What if i had to use the quotient rule...

Krizalid has a much more elegant solution (though he made a little mistake), but if you insist:

$f'(x) = \frac{4x + x^2}{(2 + x)^2}$

$f''(x) = \frac{(2 + x)^2(4 + 2x) - (4x + x^2)\left[2(2 + x)\right]}{(2 + x)^4}$

$= \frac{(2 + x)\left[(4 + 2x)(2 + x) - 2(4x + x^2)\right]}{(2 + x)^4}$ (Factoring)

$= \frac{8 + 8x + 2x^2 - 8x - 2x^2}{(2 + x)^3}$ (Reducing and expanding)

$=\frac8{(2 + x)^3}$

Which is the same result that you will get through Krizalid's method (ignoring the mistake he made).

Quote:

Originally Posted by Krizalid
There's no need to apply the quotient rule here, first let's make this thing up:

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

Now differerentiate.

Nice work, but a tiny problem:

$\frac{4x + x^2}{(2 + x)^2} = \frac{4 + 4x + x^2 - 4}{(2 + x)^2}$

$=\frac{(2 + x)^2 - 4}{(2 + x)^2} = 1 - \frac4{(2 + x)^2}$

Have you spent so much time killing integrals that you've gone soft on the derivatives? :)
• Jun 24th 2008, 04:17 PM
Krizalid
Yeah, it was a slight typo. :D
• Jun 24th 2008, 04:43 PM
snakeman11689
Eventhough, I didn't understand what Krizalid did with my problem, I thank him for his contribution to my problem, so Thank You!!!

Thanks to Reckoner, I think i understand a bit more of the complex derivative that i did, It helped so much, thanks a bunch.
• Jun 24th 2008, 05:24 PM
Reckoner
Quote:

Originally Posted by snakeman11689
Eventhough, I didn't understand what Krizalid did with my problem, I thank him for his contribution to my problem, so Thank You!!!

Krizalid used a technique called completing the square. You should have learned it in algebra, and if you have forgot it, I suggest you review as it is useful in many situations (especially with integration).

The idea is this: suppose you have some expression of the form $x^2 + bx$. This type of expression can be inconvenient since you cannot directly combine the two terms due to the different exponents. So we add a constant value to the expression in such a way that we can factor it into the form $(x + a)^2$. This is useful, for example, in solving quadratic equations (and indeed, this is where the quadratic formula comes from; see my derivation below).

Here is how to complete the square. Take your two terms $a_0x^2 + a_1x$ (ignore anything else in the expression) and factor out the $a_0$ to get the expression into the form $x^2 + bx$. Now, our goal is to get this in the form $(x + c)^2$ for some constant $c$. If you expand, we get $x^2 + 2cx + c^2$. We already have the $x^2$, and setting $c = \frac b2$ gives the linear term. Thus, we only need to add $c^2 = \left(\frac b2\right)^2$. But if we add it, we must also subtract it so that the expression remains the same:

$x^2 + bx = x^2 + bx + \left(\frac b2\right)^2 - \left(\frac b2\right)^2$

$=\left(x + \frac b2\right)^2 - \left(\frac b2\right)^2$

So, to summarize, these are the steps to complete the square:

* Start with some expression $x^2 + bx$, where $b$ can be any real number. If the $x^2$ has a coefficient other than 1, factor it out first.

* Add and subtract the square of half of $b$: $x^2 + bx + \left(\frac b2\right)^2 - \left(\frac b2\right)^2$

* Factor the first three terms: $\left(x + \frac b2\right)^2 - \left(\frac b2\right)^2$

*Done!

As an example, here is how you can use the method of completing the square to derive the quadratic formula:

Consider the general quadratic equation in $x$:

$ax^2 + bx + c = 0$

To find $x$, we will complete the square:

$ax^2 + bx + c = 0$

$\Rightarrow a\left(x^2 + \frac{bx}a + \frac ca\right) = 0$

$\Rightarrow x^2 + \frac{bx}a + \frac ca = 0$

$\Rightarrow x^2 + \frac{bx}a = -\frac ca$

$\Rightarrow x^2 + \frac{bx}a\;{\color{red} + \left(\frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2} = -\frac ca$

$\Rightarrow \left(x + \frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2 = -\frac ca$

$\Rightarrow \left(x + \frac{b}{2a}\right)^2 = \left(\frac{b}{2a}\right)^2 - \frac ca$

$\Rightarrow x + \frac{b}{2a} = \pm\sqrt{\left(\frac{b}{2a}\right)^2 - \frac ca}$

$\Rightarrow x = -\frac{b}{2a}\pm\sqrt{\frac{b^2}{4a^2} - \frac{4ac}{4a^2}}$

$\Rightarrow x = -\frac{b}{2a}\pm\sqrt{\frac{b^2 - 4ac}{4a^2}}$

$\Rightarrow x = -\frac{b}{2a}\pm\frac{\sqrt{b^2 - 4ac}}{2a}$

$\Rightarrow x = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$

Try to understand the process; it is very useful.