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differentiating lnx

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    would differentiating xlnx give 1/x?

    the lnx part gives 1/x and the x 1 so is that correct?

    thanks
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    No, it's incorrect. Are you familiar with the product rule?
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    Product rule: x(lnx)' + x'(lnx)=1+lnx

    So, what you said in yuor first line is wrong, but the second line is kind of correct...
  4. Offline

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    no.

    product rule

    differentiating a product uv = v(du/dx) + u(dv/dx)

    u = x

    v = lnx

    therefore differentiating xlnx = lnx(1) + x(1/x) = lnx + 1
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    No it won't.

    Using product rule, it becomes [ x * 1/x ] + [ 1/x ] or in other words, 1/x + 1.

    Edit: Holy ****, this forum is full of mathematical hawks.

    OP torn to pieces :eek:
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    (Original post by sdt)
    No it won't.

    Using product rule, it becomes [ x * 1/x ] + [ 1/x ] or in other words, 1/x + 1.

    Edit: Holy ****, this forum is full of mathematical hawks.

    OP torn to pieces :eek:
    don't you mean lnx? ^^

    (post farming ftw)
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    thanks all
  8. Offline

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    (Original post by amjw)
    don't you mean lnx? ^^

    (post farming ftw)

    Doh! Yes indeed. I went over this about 30 times trying to differentiate x^x. Maybe I'm mathematically challenged :rolleyes:

    Not post farming, I didn't even realise what I wrote.
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    (Original post by sdt)
    Doh! Yes indeed. I went over this about 30 times trying to differentiate x^x. Maybe I'm mathematically challenged :rolleyes:

    Not post farming, I didn't even realise what I wrote.
    i meant me
  10. Offline

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    :]
  11. Offline

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    ... over 2 year bump.
  12. Offline

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    Doh! Yes indeed. I went over this about 30 times trying to differentiate x^x. Maybe I'm mathematically challenged
    To differentiate x^x write it as:
    e^{xlnx}
    And differentiate using the product and chain rule.
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    (Original post by anshul95)
    To differentiate x^x write it as:
    e^{xlnx}
    And differentiate using the product and chain rule.
    or he can use the logarithmic differentiation
    y=x^x
    lny=x\cdot lnx
    differenciating both sides
    \frac{1}{y}y'=lnx+1
    y'=x^x(lnx+1)
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    @ztibor yes you could do that but I didn't know whether or not the person knew implicit
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    you should use Chain rule

  16. Offline

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    (Original post by safety280)
    you should use Chain rule

    You waited 6 months after the almost-three-years-old thread had died to post something which isn't even correct (or is too vague to be used in any sensible way)... :p:

    Oh, and welcome to TSR.
  17. Offline

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    (Original post by Farhan.Hanif93)
    You waited 6 months after the almost-three-years-old thread had died to post something which isn't even correct (or is too vague to be used in any sensible way)... :p:

    Oh, and welcome to TSR.

    hey, first, I was googling that function then I got a link for this thread , chain rule in this case is perfectly applied and sensible.


    thanks
  18. Offline

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    (Original post by safety280)

    hey, first, I was googling that function then I got a link for this thread , chain rule in this case is perfectly applied and sensible.


    thanks
    product rule actually.
  19. Offline

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    (Original post by anshul95)
    product rule actually.
    yeah it's actually tomato tometo things

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