1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    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


    User Avatar
    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Threads - Finding derivatives Date
How to find the derivative of this function Feb 13, 2018
Finding the min value using the derivative Feb 4, 2018
Find the derivative of this function Jan 12, 2018