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

[vanmaiden]

Homework Statement

I am not sure how to exactly solve this problem: dy/dx (x+1)^2 / (x+2)^3

Homework Equations

chain rule, product rule

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?

 

[micromass]

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}...

 

[vanmaiden]

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}...

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

 

[Mark44]

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

In English, we say, "When differentiating a product, ..."

Homework Statement

I am not sure how to exactly solve this problem: dy/dx (x+1)^2 / (x+2)^3

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}

Homework Equations

chain rule, product rule

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?

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}

 

[vanmaiden]

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.

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.

 

[Mark44]

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.