# Chain rule in division rule

1. May 6, 2007

### ETuten

1. The problem statement, all variables and given/known data

y = (2x-3)/(x^2+4)^2

2. Relevant equations

3. The attempt at a solution

I am trying to relearn the calculus that I forgot from many moons ago. I am struggling with the chain rule in the above example. I tried to set it up as follows:

This is what I know u=x^2+4 u'=2x

[(x^2+4)^2*Dx (2x-3)-(2x-3) ?? ]/(x^2+4)^4

I am confused when it comes to setting up the second half. Any help would be much obliged. I just can't seem to understand how to set up the problem

Thanks
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. May 6, 2007

### Noober

$$y=\frac{2x-3}{(x^2+4)^2}$$
$$\frac{dy}{dx}=\frac{2(x^2+4)^2-(2x-3)(2(x^2+4)(2x))}{(x^2+4)^4}$$
$$=\frac{2(x^2+4)^2-(4x^2-12x)(x^2+4)}{(x^2+4)^4}$$
$$=\frac{2(x^2+4)-(4x^2-12x)}{(x^2+4)^3}$$

How do I make a new line? \\ doesn't seem to work.

Last edited: May 6, 2007
3. May 6, 2007

### Office_Shredder

Staff Emeritus
I would imagine just breaking up the tex tags works

$$y=\frac{2x-3}{(x^2+4)^2}$$

$$\frac{dy}{dx}=\frac{2(x^2+4)^2-(2x-3)(2(x^2+4)(2x))}{(x^2+4)^4}$$

$$=\frac{2(x^2+4)^2-(4x^2-12x)(x^2+4)}{(x^2+4)^4}$$

$$=\frac{2(x^2+4)-(4x^2-12x)}{(x^2+4)^3}$$

4. May 7, 2007

### ETuten

Thanks for the reply with setting up the Calculus part. Now I know that my algebra is a little rusty, but is there a mistake in the change from the following two lines? Shouldn't it be
(8x^2-12x)(x^2+4) not (4x^2-12x)(x^2_4)

$$=\frac{2(x^2+4)^2-(4x^2-12x)(x^2+4)}{(x^2+4)^4}$$

$$=\frac{2(x^2+4)-(4x^2-12x)}{(x^2+4)^3}$$

5. May 7, 2007

### Vagrant

Yes it should be 8x^2-12x.

6. May 7, 2007