Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Stuck getting derivative when can't isolate my variable

  1. Feb 28, 2015 #1
    Hi. This is not a homework assignment. I am working to get an extrema on a graph that involves a bunch of functions and got stuck on one step:

    How to get the derivative of:
    [tex]\frac{dy}{dn} = \frac{nc(a+b)}{nc+a}[/tex]

    I can't get "n" in a place where I recognize how to get the derivative of it. I get stuck here after the first step..
    [tex] nc(a+b)*(nc+a)^{-1}[/tex]

    or do I use the quotient rule where it ends up:
    [tex] \frac{[nc(a+b)]'*(nc+a) - nc(a+b)(nc+a)'}{(nc+a)^2} [/tex]
    [tex] \frac{c(a+b)*(nc+a) - nc(a+b)(c)}{(nc+a)^2} [/tex] ??

    Can anyone please help?
    Thanks very much in advance
    Last edited: Feb 28, 2015
  2. jcsd
  3. Feb 28, 2015 #2


    Staff: Mentor

    In other words, you want ##\frac{d^2 y}{dn^2}##, right? That is, the derivative, with respect to n, of dy/dn.
    Since you are differentiating with respect to n, then the derivative of n (with respect to n) is just 1. ##\frac{d}{dn}(n) = 1##.
    You can use either the product rule (that you started with, above) or the quotient rule - your call. The product rule is usually easier to apply.
    I'm not sure I understood what you were asking. If what I wrote isn't what you're looking for, please clarify your question.
  4. Feb 28, 2015 #3
    Hi. Thanks. I'm not quite sure what you meant about the d squared y over d n squared. But it sounds like my quotient math above was correct. I wasn't sure if that was right at all.
  5. Feb 28, 2015 #4
    Actually. I think I know what you meant. I think I wrote that wrong. I meant [tex]y=\frac{nc(a+b)}{nc+a}[/tex]
  6. Feb 28, 2015 #5


    Staff: Mentor

    So all is good, right? You can use either the product rule or the quotient rule for this problem.

    I prefer the Leibniz notation for problems like this, rather than the Newton notation. IOW, I prefer d/dn[nc + a] over (nc+a)′, as it isn't clear in the latter notation that you are differentiating with respect to n.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook