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  1. #1
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    Question y'

    am i doing this right, and how to make the answer more compact:
    1: y=x-x*ln(x)
    y'= -x*(1/x) + -ln(x) = -x/x + -ln(x)

    2: y=xe^(2x)
    y'=x*2e^(2x)+e^2(x)

    3: y=3x*log(x)
    y'= 3x*1/(x*ln(10))+3*log(x) = 3x/(x*ln(10))+3*log(x) = 3/ln(10)+3*log(x)
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by stekemrt View Post
    am i doing this right, and how to make the answer more compact:
    1: y=x-x*ln(x)
    y'= -x*(1/x) + -ln(x) = -x/x + -ln(x)
    this is wrong

    use the product rule and remember you have an x in front where the derivative is 1 (and also, -x/x = -1)


    2: y=xe^(2x)
    y'=x*2e^(2x)+e^2(x)
    this is fine. to get a more compact answer, factor out the e^(2x)

    3: y=3x*log(x)
    y'= 3x*1/(x*ln(10))+3*log(x) = 3x/(x*ln(10))+3*log(x) = 3/ln(10)+3*log(x)
    this is also wrong. remember that \log_{10}x = \frac {\ln x}{\ln 10} and use the product rule
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  3. #3
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    ok,
    1: y'=1-1*ln(x)+-x*(1/x) = -ln(x)

    right?

    EDIT:
    3: y'=3*log(x)+3x*1/(x*ln(10)) = 3*log(x)+3/(ln(10))
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  4. #4
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    Quote Originally Posted by Jhevon View Post
    this is fine. to get a more compact answer, factor out the e^(2x)
    i dont know how to do that
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by stekemrt View Post
    ok,
    1: y'=1-1*ln(x)+-x*(1/x)

    right?
    you can simplify that so much more

    EDIT:
    3: y'=3*log(x)+3x*1/(x*ln(10)) = 3*log(x)+3/(ln(10))
    i have no idea what you did. let me state it explicitly, what you want is to find the derivative of:

    y = 3x \cdot \frac {\ln x}{\ln 10} = \frac 3{\ln 10}x \ln x

    and remember, ln 10 is a constant
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by stekemrt View Post
    i dont know how to do that
    e^{2x}(2x + 1)
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