How do I apply the chain rule in division rule for this calculus problem?

[ETuten]

Homework Statement

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

Homework Equations

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

 

[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.

 

[Office_Shredder]

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

I would imagine just breaking up the tex tags works

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

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

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

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

 

[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)


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

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

 

[Vagrant]

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

 

[Noober]

Bad algebra, sorry.

 

[ETuten]

well my algebra is very rusty to say the least... I have been out of math class a few years, and am working towards going back. If I was sure I was right I wouldn't have asked if it was wrong. Thanks again for all the help.