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Homework Help: Finding derivatives

  1. Jun 3, 2014 #1
    find the indicated derivative
    dp/dq if p = 1/(√q+1)

    I apologize ahead of time if you can't read my work.

    my work

    [(1/(√(q+h+1))) - (1/(√(q+1))] [itex]\div[/itex]h

    [((√(q+1)) - (√(q+h+1)))/((√(q+h+1))(√(q+1)))] [itex]\div[/itex]h

    [(q+1-q-h-1)/(((√q+h+1)(√q+1))((√q+1)+(√q+h+1)))][itex]\div[/itex] h

    [-h//(((√(q+h+1))(√(q+1)))((√(q+1))+(√(q+h+1))))][itex]\div[/itex] h



    -1/[(q+1)(2√(q+1))] this was my answer

    the answer in the book is -1/[2(q+1)(√(q+1))]

    is my answer the same as the book or is there something else I still need to do?
    Last edited: Jun 3, 2014
  2. jcsd
  3. Jun 3, 2014 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

    Do you mean
    [tex] p = \frac{1}{\sqrt{q+1}} \text{ or } p = \frac{1}{\sqrt{q} + 1}?[/tex]
    In text you should write the first as p = 1/√(q+1) and the second as p = 1/(1+√p) or 1/((√p)+1).
  4. Jun 3, 2014 #3
    I am sorry it is the first one [tex] p = \frac{1}{\sqrt{q+1}} [/tex]
  5. Jun 3, 2014 #4


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    Gold Member

    So your answer was $$-\frac{1}{(q+1)(2\sqrt{q+1})}$$ and the book's was $$-\frac{1}{2(q+1)(\sqrt{q+1})}?$$

    If so, yes they are the same. q is just a variable. It is the norm to see, for example, 2x rather than x2. And the latter becomes problematic when dealing with a product of different variables.
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