Math Help - Finding Derivative by Quotient Rule

1. 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.

2. 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.

3. 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...

4. 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).

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?

5. Yeah, it was a slight typo.

6. 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.

7. 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.