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Math Help - Derivative of Logarithmic Function

  1. #1
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    Derivative of Logarithmic Function

    How do I derive y = ln(xe^x^3)? Do I do it like this:

    Use the formula:
    1/(xe^x^3)

    Multiply it times the derivative, found using the product rule:
    1/(xe^x^3) * (xe^x^3) + e^x^3

    Simplify:
    ((xe^x^3) + e^x^3) / (xe^x^3)

    Divide out e^x^3:
    (x + 1)/ x

    Sorry; I don't know how to make the calculations look good when typing. So, is that right? I have the feeling I did something wrong, so that's why I'm asking. Thanks!
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  2. #2
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    Quote Originally Posted by Maziana View Post
    How do I derive y = ln(xe^x^3)?
    use the laws of logs first ...

    y = \ln(x \cdot e^x \cdot x^3)

    y = \ln(x) + \ln(e^x) + \ln(x^3)

    y = \ln(x) + x + 3\ln(x)

    y = 4\ln(x) + x

    y' = \frac{4}{x} + 1
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  3. #3
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    I think I wasn't clear because of my bad formatting...

    the problem is not y = ln((x) (e^x) (x^3))

    it is y = ((x) (e^x^3)), so e is raised to the xth power, which is then raised to the 3rd power.

    Unless you mean I can somehow break it down like that...?
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  4. #4
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    Quote Originally Posted by Maziana View Post
    I think I wasn't clear because of my bad formatting...

    the problem is not y = ln((x) (e^x) (x^3))

    it is y = ((x) (e^x^3)), ??? so e is raised to the xth power, which is then raised to the 3rd power.
    is it

    y = \ln(x \cdot e^{x^3})

    or is it just

    y = x \cdot e^{x^3}

    ???
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  5. #5
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    yeah, the first one... sorry!
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  6. #6
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    y = \ln(x \cdot e^{x^3})

    once again, use the laws of logs first ...

    y = \ln{x} + \ln(e^{x^3})

    y = \ln{x} + x^3

    y' = \frac{1}{x} + 3x^2
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  7. #7
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    Hello, Maziana!

    Differentiate: . y \:=\: \ln\left(xe^{x^3}\right)
    I'll do it two ways . . .


    [1] Chain Rule and Product Rule.

    . . y \;=\;\ln\left(x\cdot e^{x^3}\right)

    . . \frac{dy}{dx} \;=\;\frac{1}{x\cdot e^{x^3}}\cdot\left(x\cdot3x^2\cdot e^{x^3} + 1\cdot e^{x^3}\right) \;=\;\frac{1}{x\cdot e^{x^3}}\cdot e^{x^3}\left(3x^3+1\right) . = \;\frac{3x^3+1}{x}


    [2] Simplify first.

    . . y \;=\;\ln\left(x\cdot e^{x^3}\right) \;=\;\ln(x) + \ln\left(e^{x^3}\right) \;=\;\ln(x) + x^3\cdot\ln(e) \;=\;\ln(x) + x^3

    . . \frac{dy}{dx} \;=\;\frac{1}{x} + 3x^2 \;=\;\frac{1+3x^3}{x}

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