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Finding derivative of y + 1/(y+3)

  1. Oct 26, 2005 #1
    Hi, I am stuck on this question. Can someone please help me as quick as possible, I am revising for an exam on Monday.

    If A= y + 1/(y+3) , find dA/dy.

    I have been tryin to solve this question but I keep getting the wrong answer.:confused: :confused:
  2. jcsd
  3. Oct 26, 2005 #2


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    What have you tried so far?
  4. Oct 26, 2005 #3


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    I assume you know that differentiation is lineair? So (f+g)' = f' + g'.
    You should at least be able to do a part then :smile:
  5. Oct 26, 2005 #4
    Well I tried to make the equation simpler first. y + 1/(y+3) = (y^2 + 3y + 1)(y+3)^-1 right so far?
  6. Oct 26, 2005 #5
    Sorry, confused. I just don't know how to start the question in the first place.
  7. Oct 26, 2005 #6


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    That's right but it doesn't make it simpler!
    You're probably used to factoring a lot in order to solve equations etc, but when you're going to differentiate (or integrate, later) you'll see that it's easier to differentiate sums than products!

    Now, if you need to find the derivative of a sum, you can just take the derivative of each term. In your case (a ' denotes a derivative):

    [tex]\left( {y + \frac{1}{{y + 3}}} \right)^\prime = \left( y \right)^\prime + \left( {\frac{1}{{y + 3}}} \right)^\prime [/tex]
  8. Oct 26, 2005 #7
    That gives me dA/dy= 1 - 1/(y+3)^2. Is that right?
  9. Oct 26, 2005 #8


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    That is correct :smile:
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