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I seem to lack an implicit differentiation technique, should be a quick fix

  1. May 28, 2012 #1
    1. The problem statement, all variables and given/known data

    [itex]u = x^u + u^y[/itex]

    Find the partial derivatives of ##u## w.r.t. ##x, y##.


    2. Relevant equations

    Only the one.

    3. The attempt at a solution

    I've attempted reducing the problem using logs, but the resulting equations seem no more tenable to me. I'm sure there is a nice trick... it isn't currently in my tool belt.


    Edit: A further request if anybody has this information at their fingertips. I've been two years removed from my studies of mathematics and physics... I am planning to return as I've found the 'real world' much less interesting and inspiring. I've been going through some old textbooks as a warmup, and although I can still recall many of the intricate mathematical methods used in various branches of physics, my knowledge of elementary manipulations is pretty darn tattered. Mary Boas' text in mathematical methods in physics has been a *great* boon; I am curious as to whether or not a similar text exists for one who desires a "handbook" of mathematical theorems, identities, proofs &c. without all of the detailed explanations and problem sets.
     
    Last edited: May 28, 2012
  2. jcsd
  3. May 28, 2012 #2
    I think it boils down to the following: You find it hard to differentiate functions of the form

    [tex]f(x) = g(x)^{h(x)}\text.[/tex]

    I think it helps to first work out a general formula for this, i.e. something kind of like the product rule, but for powers. Then, you can go back and tackle your original question.

    To find such a formula, it helps to (as you tried on your original problem) take logs on both sides.
     
  4. May 28, 2012 #3
    Thanks, Ayre!

    Seems to boil down to a combination of the ubiquitous chain rule for logarithms of functions and the product rule. So clear now! I should've seen it... hopefully a bit of mathematical intuition returns with study.
     
  5. May 28, 2012 #4
    [tex]
    \begin{cases}
    \frac{\partial u}{\partial x}=\frac{u}{x(1-u\ln{x})}+\frac{u\ln{u}}{1-y}\frac{\partial y}{\partial x}\\

    \frac{\partial u}{\partial y}=\frac{u\ln{x}}{1-y}+\frac{u^2}{x(1-u\ln{x})}\frac{\partial x}{\partial y}
    \end{cases}
    [/tex]

    is what I've got so far.

    I began doing a bunch of algebra, but figured I'd ask if there were some theorem that might let me solve this with a simple determinant or something instead.
     
  6. May 28, 2012 #5

    Curious3141

    User Avatar
    Homework Helper

    That doesn't look right. When doing the partial diff by x, treat y as a constant, and vice versa. The terms [itex]\frac{\partial y}{\partial x}[/itex] and [itex]\frac{\partial x}{\partial y}[/itex] make no sense.

    It might help, if in your working, you replaced the other variable with "k" (which your brain "recognises" as a constant), and just used the usual derivative symbol e.g. [itex]\frac{du}{dx}[/itex]. Then work through everything, and put the other symbol back when you present your final answer.
     
    Last edited: May 28, 2012
  7. May 29, 2012 #6
    ##y## isn't a constant with respect to ##x##. The original equation defines a function implicitly; the value of any parameter is dependent upon the other two. I've plugged the equation into mathematica, and the ##\frac{\partial y}{\partial x}## and ##\frac{\partial x}{\partial y}## terms which I've worked out on my own, combined with my partial solution above, contain all of the quantities displayed in the mathematica solution. I'm pretty sure that I'm on the right track. My problem is that the only way I can think of to solve is with horribly tedious algebraic manipulations of the differentials, all of which are very nested within each other.

    I was hoping there might be some analog to Cramer's rule for situations such as this.
     
  8. May 29, 2012 #7
    Let's look at the partial derivative w.r.t. x. You ought to treat y as a constant, and u(x)[/i] as a function of x only, in all the calculations.

    Let us look at the first term on the right-hand side: [itex]x^{u(x)} = \exp \left( u(x) \, \ln x \right)[/itex]. Apply the "chain rule" and the product rule to get:
    [tex]
    \frac{\partial}{\partial x} \left[ \exp \left( u(x) \, \ln x \right) \right] = \exp \left( u(x) \, \ln x \right) \, \left( u'(x) \, \ln x + u(x) \, \frac{1}{x} \right)
    [/tex]
    Here [itex]u'(x) = \partial u/\partial x[/itex].

    Next, let us look at the second term [itex]\left[ u(y)\right]^{y}[/itex]. This is a power function. Apply the chain rule and the power rule to get:
    [tex]
    \frac{\partial}{\partial x} \left[ u(x) \right]^{y} = y \, \left[ u(x) \right]^{y - 1} \, u'(x)
    [/tex]
    The derivative on the left-hand side is obvious. Collect all the terms with [itex]u'(x)[/itex] and solve for this (as an algebraic equation). This is your partial derivative obtained implicitly.
     
  9. May 29, 2012 #8

    Curious3141

    User Avatar
    Homework Helper

    I never said that. I said that when working out the partial derivative of ##u## wrt ##x##, you have to treat ##y## as (if it were) a constant. Similarly, when working out the partial derivative of ##u## wrt ##y##, treat ##x## as (if it were) a constant.

    Sure. But when you take the partial derivative of one variable wrt one other variable, you have to treat the other variable(s) like constant(s).

    If you were asked to determine [itex]\frac{\partial y}{\partial x}[/itex], for instance, u would be treated like a constant.

    Mathematica will output exactly what you ask it for. Did you actually ask it to work out [itex]\frac{\partial u}{\partial x}[/itex] and [itex]\frac{\partial u}{\partial y}[/itex]? Because those should NOT have terms like [itex]\frac{\partial y}{\partial x}[/itex] or [itex]\frac{\partial x}{\partial y}[/itex] in them.

    I don't think so. Please look at Dickfore's post as well.
     
  10. May 29, 2012 #9
    Try using the functional power rule.

    $$ (f^g)' = \left(e^{g\ln f}\right)' = f^g\left(f'{g \over f} + g'\ln f\right),\quad $$
     
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