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

  1. Jun 16, 2011 #1
    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. jcsd
  3. Jun 16, 2011 #2

    micromass

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    Staff Emeritus
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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]...
     
  4. Jun 16, 2011 #3
    Ah, I see. Thank you very much micromass for clearing that up for me.
     
  5. Jun 16, 2011 #4

    Mark44

    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
    [tex]\frac{d}{dx}\frac{(x + 1)^2}{(x + 2)^3}[/tex]
    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]
     
  6. Jun 17, 2011 #5
    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.
     
  7. Jun 17, 2011 #6

    Mark44

    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.
     
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