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Purpose of differentiating both sides

  1. Jan 7, 2009 #1
    Hey guys,
    I've just recently started calculus, and just trying to get to grips with it.
    I'll use an example for my question; if [tex]y=e^x[/tex], then the source suggests rearranging to [tex]\ln y=x[/tex], the differentiating with respect to x yields [tex]\frac{\delta}{\delta{x}} (ln\ y)=1[/tex].

    I understand that for my original function I wouldn't be able to find [tex]\delta{y}[/tex], so it's rearranged, but what are the consequences of this? As it appears that the requirement is to DIFFERENTIATE BOTH SIDES. Do I always do this when the equation has been altered? I can't piece togeather why this is happening. Hopefully someone can help.
    Thanks.
    Nobahar.
     
  2. jcsd
  3. Jan 7, 2009 #2

    rock.freak667

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    [tex]y=e^x \Rightarrow lny=x[/tex]


    [tex]\frac{d}{dx}(lny)=\frac{d}{dx}(x)[/tex]

    [tex]= \frac{dy}{dx}\frac{d}{dy}(lny)=1[/tex]

    I didn't fully understand what you were asking so I tried my best to alleviate your confusion.
     
  4. Jan 7, 2009 #3

    Dick

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    As to the purpose, if you continue with this, since y=e^x then dy/dx=y. So you've shown y*d(ln(y))/dy=1. So d(ln(y))/dy=1/y. You've used the fact you know how to differentiate the exp function to figure out how to differentiate it's inverse function, ln.
     
  5. Jan 12, 2009 #4
    Cheers guys! :smile:
     
  6. Jan 12, 2009 #5

    Mark44

    Staff: Mentor

    Other folks showed you what to do, but I don't think they answered your questions, so I'll take a shot at it.
    Your goal here was to differentiate e^x, but didn't have a formula to do so.
    What you called "rearranging" the equation is to rewrite the original equation, where y is a function of x, as a new equation where x is a different function of y. In fact, the different function is the inverse of the exponential function.

    Apparently you already know the derivative of ln(x), so using the chain rule, you can differentiate ln(y) with respect to x (i.e., find d/dx(ln(y)), which as Dick pointed out is 1/y * dy/dx.
    It doesn't have anything to do with whether an equation has been altered. Whenever you differentiate one side of an equation, you have to differentiate the other side as well. And the differentiation has to be with respect to the same variable. IOW, if you differentiate one side with respect to x, you have to differentiate the other side with respect to x as well.
     
  7. Jan 14, 2009 #6
    That's pretty much answered my questions fully!
    Thanks!
     
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