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Homework Help: Derivative of f(x) to the power of g(x), and algebra problem

  1. Nov 19, 2008 #1
    1. The problem statement, all variables and given/known data

    1. If f(x), g(x) and h(x) are real functions of x, show that

    when [tex]h(x)=[f(x)]^{g(x)}[/tex]

    then [tex]h'(x)=[f(x)]^{g(x)}(g'(x)\ln[f(x)]+g(x)\frac{f'(x)}{f(x)})[/tex]

    2. [tex]\frac{A}{x}+\frac{B}{P-x}=\frac{1}{x(P-x)}[/tex] where x is a variable, and P is a constant. Find A and B in terms of P.

    2. Relevant equations

    3. The attempt at a solution

    1. I start by doing what I usually do, like with [tex]x^{x^2}[/tex]:


    Looking at the derivative, I see


    Which looks nothing like what I got :(

    2. Getting a common denominator and canceling:


    Then by inspection,


    It was a fluke that I got that :/. So I'm wondering how to prove it arithmetically, or just some general method of solving these kinds of problems for when I come across them again.
  2. jcsd
  3. Nov 19, 2008 #2
    Hi Deadleg,

    If we can get your example [tex]x^{x^2}[/tex] sorted then you will be fine to do the question at hand. We rewrite it as [tex]x^{x^2} = e^{x^2\log{x}[/tex], and then differentiate (using the chain rule), obtaining [tex](2x\log{x} + x})e^{x^2\log{x}[/tex], i.e., [tex](2x\log{x} + x})x^{x^2}[/tex].

    The exact same technique applies to your problem.

    Actually, you're not quite right (did you test your inspection?). From your (correct) equation A(P-x)+Bx=1, we can then gather like terms: x(-A + B) + AP = 1, and then equate coefficients (of powers of x).
  4. Nov 19, 2008 #3


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    You can't treat (f(x))^(g(x)) like it was a power function like x^n. f(x)^g(x)=e^(log(f(x)*g(x)). Use the chain rule on that. For the second one, you've got 1=AP-Ax+Bx=AP+(B-A)x. Since those are supposed to be equal for ALL values of x, you must have AP=1 since that's the constant the right side, and (B-A)=0 since there is no x on the right side. And I don't think you got it right.
  5. Nov 20, 2008 #4
    Ah, so for 1.




    And for 2.



    [tex]AP=1,\ A=\frac{1}{P}[/tex]

    [tex]B-A=0,\ B=A=\frac{1}{P}[/tex]
  6. Nov 20, 2008 #5


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    That's it exactly.
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