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CALCULUS! Easy for you maybe, impossible to me...

Discussion in 'Off-Topic' started by SomeCanadianGuy, Mar 24, 2004.

  1. SomeCanadianGuy

    SomeCanadianGuy IncGamers Member

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    CALCULUS! Easy for you maybe, impossible to me...

    I really, really need help with this. It's a simple enough question, but I can't seem to figure it out...

    What's the differential of the function y = xln(x) ?

    I've got the answer. The problem is that I don't understand how to get to the answer.... I'm lost somewhere in the path in between, and it's annoying the hell out of me. Any help out there would be immensely appreciated. I'll even hire Anakha to give a lapdance to the one who succesfully explains it to me. :D
     
  2. cleanupguy

    cleanupguy IncGamers Member

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    Isn't the answer y' = x'ln(x) + x*1/x = x'ln(x) +1 ?

    I'm not sure if there is anything to work out since I just did this by looking at the problem. Are you asking why d/dx of ln(x) is 1/x?

     
  3. SomeCanadianGuy

    SomeCanadianGuy IncGamers Member

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    The answer given to me is: dy = (1 + lnx)dx

    Pourquoi? I don't know...
     
  4. cleanupguy

    cleanupguy IncGamers Member

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    Well, I was sloppy with the answer a bit.

    My answer was that y' = x'ln(x) + 1..

    This is : dy/dx = 1 + ln (x)
    (Divide both sides by dx and you get the answer above)
    dy/dx is y'

    Are you asking why the answer is like that or are you asking why the derivative of ln(x) is (1/x)dx? If you specify, I might be able to show work since I'm not so busy at the moment.

     
  5. SomeCanadianGuy

    SomeCanadianGuy IncGamers Member

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    Thank you very much! Uhm... Right, my question would be why exactly does it become 1 + ln(x)? I know that the derivative of ln(x)=(1/x) but I can't figure out how or why this step x'ln(x) + x*1/x comes into existence... It just doesn't seem to work out in my head.
     
  6. KawaiiWolf

    KawaiiWolf IncGamers Member

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    It's the Chain Rule.

    Take this for example:
    F(x) = g(x)h(x)
    F'(x) = g'(x)h(x) + g(x)h'(x)

    In your case g(x) = x and h(x) = ln(x)
    y = xln(x)
    y' = (x)'ln(x) + x(lnx)'
    y' = ln(x) + x(1/x)
    y' = ln(x) + 1
    dy = (ln(x) + 1)dx

    Make sense?
     
  7. cleanupguy

    cleanupguy IncGamers Member

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    When you are taking a derivative of two factors, such as x * ln(x) or x * sin (x), you need to do the following.

    For example, let's say that you have something like Y = A * B. Granted that neither A nor B are constants, you will need to take a derivative of A first, multiplied by B. Added to the result is the derivative of B, multiplied by A.

    Looking at the question you had: y = x * ln(x) where x represents A and ln (x) represents B

    dy = dx * ln (x) + d (ln (x)) * x ============== (1)

    You know the derivative of x is dx and derivative of ln (x) is (1/x)dx. (Remember that, in actuality, when you are taking a derivative of x, the result is not 1 but dx.)

    Solving (1) above, you get:

    dy = dx * ln (x) + (1/x)dx * x ================ (2)

    Now, factoring out dx will give you:

    dy = dx (ln (x) + (1/x)*x) ================ (3)

    Simplifying (3) will give you:

    dy = dx (ln (x) + 1) ================= (4)

    If you even go a step farther, you will get

    dy/dx = ln (x) + 1


    I hope this helped.

     
  8. mudvaynefan

    mudvaynefan IncGamers Member

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    god... im still on alg1!

    i dont think i will ever learn that stuff
     
  9. SomeCanadianGuy

    SomeCanadianGuy IncGamers Member

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    MY GOD!!! Bloody brilliant!!! Now I just wonder why I was never taught that when you want the derivative of a multiplication of two functions (more or less), you have to take the derivative of the first multiplied by the second function, and then repeat with the second function. :rant: I've got some Cal teachers to go give a stern talking to, I believe....

    THANK YOU GUYS SOOOOOO MUCH!!! *whistles over Anakha* Yo, beefy brotha, lapdance for the gents please! :teeth:
     
  10. Underseer

    Underseer IncGamers Member

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    Waaaaaah! I want to be able to use named entities on this board! Grah!

    ∂y/∂x

    Actually, the equations in this thread were relatively painless without named entities, but I've seen other threads get really ugly!
     

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