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Math Help - calc help on integrals/differentiation

  1. #1
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    calc help on integrals/differentiation

    can anyone just double check these for me please??

    differentiation

    ln (radical((x^2) +4)) = x / (x^2) + 4

    ln (5-x)^6 = -6/(5^-x)

    ln ( x * (e^x) ) = (e^-x)((e^x)x + e^x )) / x

    ln (e^x) / radical(e^x) + 1 = 1 / (radical(e^x) + 1) - radical(e^x) log (e^x) / 2(radical(e^x) +1)^2
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  2. #2
    Junior Member
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    calc help on integrals/differentiation

    can anyone just double check these for me please??

    differentiation

    ln (radical((x^2) +4)) = x / (x^2) + 4

    ln (5-x)^6 = -6/(5^-x)

    ln ( x * (e^x) ) = (e^-x)((e^x)x + e^x )) / x

    ln (e^x) / radical(e^x) + 1 = 1 / (radical(e^x) + 1) - radical(e^x) log (e^x) / 2(radical(e^x) +1)^2
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  3. #3
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    Hello, darkblue!

    If no one is responding, it's because your work is hard to read.


    . . . . . . - - - . ______
    1. f(x) .= .ln(√x + 4)
    . . . . . . - - - . ._____
    . . f'(x) .= .x /√x + 4 . Right!


    2. f(x) .= .(5 - x)^6

    . . f'(x) .= .-6/(5 - x) . Yes!


    3. f(x) .= .ln(xe^x)

    . . f'(x) .= .(e^-x)(xe^x + e^x ) / x . ??
    Where did the e^{-x} come from?

    We have: .f(x) .= .ln(xe^x) .= .ln(x) + ln(e^x) .= .ln(x) + x

    . . Therefore: .f'(x) .= .1/x + 1 .= .(1 + x)/x



    4. f(x) = ln (e^x) / radical(e^x) + 1
    . . f'(x) = 1 / (radical(e^x) + 1) - radical(e^x) log (e^x) / 2(radical(e^x) +1)^2 . ??
    Did you know that: ln(e^x) = 1 ?

    . . . . . . . . . . . . . . . . . . . . . . 1
    If the function is: .f(x) .= . ------------ .= .(e^x + 1)^{-}
    . . . . . . . . . . . . . . . . . . . √e^x + 1

    . . the derivative is: .f'(x) .= .(-)(e^x)(e^x + 1)^{-3/2}

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  4. #4
    Math Engineering Student
    Krizalid's Avatar
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    Remember that



    If you apply this, with the first one, you've got:



    Do the same with the others.

    P.S.: this forum hasn't LaTeX?
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Krizalid View Post
    P.S.: this forum hasn't LaTeX?
    It's a work in progress...

    -Dan
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  6. #6
    Math Engineering Student
    Krizalid's Avatar
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    Great!

    Thanks for the notice.
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