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

  • Thread starter vanmaiden
  • Start date
  • #1
102
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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?
 
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Answers and Replies

  • #2
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Hi vanmaiden! :smile:

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

[tex](fg)^\prime=f^\prime g+fg^\prime[/tex]

Now [itex](x+1)^2(x+2)^{-3}[/itex] is a product with [itex]f(x)=(x+1)^2[/itex] and [itex]g(x)=(x+2)^{-3}[/itex]...
 
  • #3
102
1
Hi vanmaiden! :smile:

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

[tex](fg)^\prime=f^\prime g+fg^\prime[/tex]

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

Ah, I see. Thank you very much micromass for clearing that up for me.
 
  • #4
35,294
7,152
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
[tex]\frac{d}{dx}\frac{(x + 1)^2}{(x + 2)^3}[/tex]

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:
[tex]\frac{d}{dx}\frac{u}{v} = \frac{v u' - u v'}{v^2}[/tex]
 
  • #5
102
1
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.
 
  • #6
35,294
7,152
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.
 

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