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Math Help - Cannot find my own mistake (derivation)

  1. #1
    Member Pranas's Avatar
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    Cannot find my own mistake (derivation)

    Hello.

    So I have this:
    x^{y} = y^{x}
    which, in case x and y are positive, may be graphed like:


    Assuming:
    x^{y} = y^{x} \Longrightarrow e^{y \cdot ln(x)} = e^{x \cdot ln(y)} \Longrightarrow y \cdot ln(x) = x \cdot ln(y) \Longrightarrow ln(x^{y}) = ln(y^{x})
    It seems plotting x^{y} = y^{x} and ln(x^{y}) = ln(y^{x}) should be essentially the same.

    However, my real problem is when I try to get \frac{dy}{dx} out of those equations.

    *Unfortunately first pictures showing my thorough work had to be removed because of some reader-disruptive flaws.*

    Essence of derivation of equation #1:
    \frac{d}{dx}(x^{y} = y^{x})
    y \cdot x^{y-1} \cdot \frac{dx}{dx} + x^{y} \cdot ln(x) \cdot \frac{dy}{dx} = x \cdot y^{x-1} \cdot \frac{dy}{dx} + y^{x} \cdot ln(y) \cdot \frac{dx}{dx}
    \frac{dy}{dx} = \frac{y^{x} \cdot ln(y) - y \cdot x^{y-1}}{x^{y} \cdot ln(x) - x \cdot y^{x-1}}

    Essence of derivation of equation #2:
    \frac{d}{dx}(ln(x^{y}) = ln(y^{x}))
    \frac{y}{x} \cdot \frac{dx}{dx} + ln(x) \cdot \frac{dy}{dx} = \frac{x}{y} \cdot \frac{dy}{dx} + ln(y) \cdot \frac{dx}{dx}
    \frac{dy}{dx} = \frac{ln(y) - \frac{y}{x}}{ln(x) - \frac{x}{y}}

    Why is there a big difference?
    Last edited by Pranas; November 22nd 2010 at 10:09 AM.
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  2. #2
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    Quote Originally Posted by Pranas View Post
    Hello.

    So I was looking for the derivative of this:



    What is that very first line? This is false big time: the left hand is yx^{y-1} , whereas the right one is y^x\ln y ...

    Tonio


    But then I tried to put everything inside logarithm ln() , which, in case numbers are positive, I thought wouldn't change the answer. However it changed big time:


    Can you explain, what is wrong with my thoughts?
    .
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  3. #3
    Member Pranas's Avatar
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    tonio, could you please be at least a little more thorough?

    x^{y} = y^{x} is what i have.
    \frac{dy}{dx} = ? is what I need.

    As long as I understand you saying y \cdot x^{y-1} = y^{x} \cdot ln(y) that doesn't give me much.
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    Quote Originally Posted by Pranas View Post
    tonio, could you please be at least a little more thorough?

    x^{y} = y^{x} is what i have.
    \frac{dy}{dx} = ? is what I need.

    As long as I understand you saying y \cdot x^{y-1} = y^{x} \cdot ln(y) that doesn't give me much.

    Pranas, could you please be a little less sloppy? You did not say anything about having x^y=y^x . It doesn't

    appear anywhere in your message. It only appears the first equality \frac{d}{dx}(x^y)=\frac{d}{x}(y^x) , which gives

    you what I wrote you.

    And don't bother in writing back asking for my help unless you first apologize.

    Tonio
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  5. #5
    Member Pranas's Avatar
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    Quote Originally Posted by tonio View Post
    Pranas, could you please be a little less sloppy? You did not say anything about having x^y=y^x . It doesn't

    appear anywhere in your message. It only appears the first equality \frac{d}{dx}(x^y)=\frac{d}{x}(y^x) , which gives

    you what I wrote you.

    And don't bother in writing back asking for my help unless you first apologize.

    Tonio
    Yes, I apologize. I simply couldn't figure your post out at first, although it is very correct
    Seems I should have written \frac{d}{dx}(x^{y} = y^{x}) ?

    However x^{y} = y^{x} \Longrightarrow e^{y \cdot ln(x)} = e^{x \cdot ln(y)} \Longrightarrow y \cdot ln(x) = x \cdot ln(y) \Longrightarrow ln(x^{y}) = ln(y^{x}) so it seems like the disagreement I get
    \frac{d}{dx}(x^{y} = y^{x}) \neq \frac{d}{dx}(ln(x^{y}) = ln(y^{x})) prevails although it (I believe) really means the same.
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  6. #6
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    Quote Originally Posted by Pranas View Post
    Yes, I apologize. I simply couldn't figure your post out at first, although it is very correct
    Seems I should have written \frac{d}{dx}(x^{y} = y^{x}) ?

    However x^{y} = y^{x} \Longrightarrow e^{y \cdot ln(x)} = e^{x \cdot ln(y)} \Longrightarrow y \cdot ln(x) = x \cdot ln(y) \Longrightarrow ln(x^{y}) = ln(y^{x}) so it seems like the disagreement I get
    \frac{d}{dx}(x^{y} = y^{x}) \neq \frac{d}{dx}(ln(x^{y}) = ln(y^{x})) prevails although it (I believe) really means the same.
    Well, that last statement is certainly true- \frac{d(x^y= y^x)}{dx}\ne \frac{d ln(x^y)= ln(y^x)}{dx} and no one has said they are equal.

    The problem goes back to your initial post: "I was looking for the derivative of this
    \frac{d x^y}{dx}= \frac{d y^x}{dx}
    which is very ambiguous: were you asking for the derivative of each side of that equation or were you asking how to arrive at that equation?

    Now you say "Seems I should have written \frac{d}{dx}(x^{y} = y^{x})
    If we let z= x^y, then ln(z)= y ln(x) and so \frac{1}{z}\frac{dz}{dx}= \frac{y}{x}+ ln(x)\frac{dy}{dx}.

    That is, \frac{dz}{dx}= \frac{d x^y}{dx}= \left(\frac{y}{x}+ ln(x)\frac{dy}{dx}\right)z
    \frac{dx^y}{dx}=   yx^{y-1}+ y^x ln(x)\frac{dy}{dx}.

    Of course, that depends upon dy/dx. We cannot differentiate some function of x without knowing precisely what that function is- what y is as a function of x.

    (Note that if y does NOT depend upon x, if y is a constant, then dy/dx= 0 and this just becomes the standard \frac{dx^y}{dx}= yx^{y-1} from Calculus I.)
    Last edited by HallsofIvy; November 23rd 2010 at 04:32 AM.
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  7. #7
    Member Pranas's Avatar
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    Indeed there were some obvious flaws in my "questionnaire", I am trying to correct most of it as the conversation evolves.

    Quote Originally Posted by HallsofIvy View Post
    Well, that last statement is certainly true- \frac{d(x^y= y^x)}{dx}\ne \frac{d ln(x^y)= ln(y^x)}{dx} and no one has said they are equal...
    For the sake of simplicity, let's say we're interested only in positive values of x and y (I've added a possible graphing below in this post).
    Please be aware that I did not add ln() just like that, I tried to rationally generate it as showed in this thread:
    x^{y} = y^{x} \Longrightarrow e^{y \cdot ln(x)} = e^{x \cdot ln(y)} \Longrightarrow y \cdot ln(x) = x \cdot ln(y) \Longrightarrow ln(x^{y}) = ln(y^{x})
    Maybe I am wrong, but at this moment I do not see how were the relations between x and y effected by that, therefore I assume I haven't changed the possible plot nor the value of \frac{dy}{dx}.


    Quote Originally Posted by HallsofIvy View Post
    ...The problem goes back to your initial post: "I was looking for the derivative of this
    \frac{d x^y}{dx}= \frac{d y^x}{dx}
    which is very ambiguous: were you asking for the derivative of each side of that equation or were you asking how to arrive at that equation?...
    I did not write what I really meant to at that point. Sorry.


    Quote Originally Posted by HallsofIvy View Post
    ...Now you say "Seems I should have written \frac{d}{dx}(x^{y} = y^{x})
    If we let z= x^y, then [tex]ln(z)= y ln(x) and so \frac{1}{z}\frac{dz}{dx}= \frac{y}{x}+ ln(x)\frac{dy}{dx}.

    That is, \frac{dz}{dx}= \frac{d x^y}{dx}= \left(\frac{y}{x}+ ln(x)\frac{dy}{dx}\right)z
    \frac{dx^y}{dx}=   yx^{y-1}+ y^x ln(x)\frac{dy}{dx}...
    Well, yes. I pretty much applied parallel method to yours on both x^{y} and y^{x}. Then expressed \frac{dy}{dx} (that would be my equation #1 in updated first message). As far as my mental calculation goes, answer seems to be identical as the one you're approaching in this part of the post.


    Quote Originally Posted by HallsofIvy View Post
    ...Of course, that depends upon dy/dx. We cannot differentiate some function of x without knowing precisely what that function is- what y is as a function of x.

    (Note that if y does NOT depend upon x, if y is a constant, then dy/dx= 0 and this just becomes the standard \frac{dx^y}{dx}= yx^{y-1} from Calculus I.)
    Indeed you're correct once more. That is a reasonable variation in a broad sense, although might be a little vulgar mapped on a plane, because I would imagine it as of having only a few points.


    What I imagined as a relation between positive x and y in the function x^{y} = y^{x} is like this:

    In my language that is defined to be a simple function, only "unexpressed", because of not being represented by y = *operations involving only variable x and constants*


    P.S. We've had some confusion here, so I can politely remind, that what's still unclear for me is inequality \frac{d(x^{y}= y^{x})}{dx}\ne \frac{d(ln(x^{y})= ln(y^{x}))}{dx}.
    Also I tried to do my best in editing the first post.
    Last edited by Pranas; November 22nd 2010 at 10:09 AM.
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