# Derivative of (x+1)^2 / (x+2)^3

1. Jun 16, 2011

### vanmaiden

1. The problem statement, all variables and given/known data
I am not sure how to exactly solve this problem: dy/dx (x+1)^2 / (x+2)^3

2. Relevant equations
chain rule, product rule

3. The attempt at a solution
(x+1)^2 / (x+2)^3

(x+1)^2 * (x+2)^-3

from this point, I am unsure if I should use the chain rule and turn it into:

(2 * (x+1) * 1) * (-3 * (x+2)^-4 * 1)

should I then simplify and apply the product rule?

Last edited: Jun 16, 2011
2. Jun 16, 2011

### micromass

Staff Emeritus
Hi vanmaiden!

When deriving a product, you must always use the product rule

$$(fg)^\prime=f^\prime g+fg^\prime$$

Now $(x+1)^2(x+2)^{-3}$ is a product with $f(x)=(x+1)^2$ and $g(x)=(x+2)^{-3}$...

3. Jun 16, 2011

### vanmaiden

Ah, I see. Thank you very much micromass for clearing that up for me.

4. Jun 16, 2011

### Staff: Mentor

In English, we say, "When differentiating a product, ..."
Taking this literally, there is no problem. dy/dx is the derivative of some function y with respect to x, so the expression above is the product of an unknown derivative and the rational function (x + 1)^2 / (x + 2)^3.

It's clear that it should be
$$\frac{d}{dx}\frac{(x + 1)^2}{(x + 2)^3}$$
Certainly, you can turn a quotient into a product, and then use the product rule to find the derivative, but you can also use the quotient rule, which looks like this:
$$\frac{d}{dx}\frac{u}{v} = \frac{v u' - u v'}{v^2}$$

5. Jun 17, 2011

### vanmaiden

Yes, thank you for catching me. I see how my mistake could have been interpreted incorrectly. I was taking the derivative of (x + 1)^2 / (x + 2)^3 and not y. On a side note, how do you type with those large italics Mark44? I see it used all the time when equations are typed.

6. Jun 17, 2011

### Staff: Mentor

I was using LaTeX. Here's what I had.
[ tex ]\frac{d}{dx}\frac{u}{v} = \frac{v u' - u v'}{v^2}[ /tex ]

I added extra spaces in the tex tags, otherwise the browser will render it, and you wouldn't be able to see the LaTeX script I used.