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Math Help - differentiate with respect to x

  1. #1
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    Unhappy differentiate with respect to x


    (a) y = e^x ln x + x^3
    Hint: Note that the first term on the right hand side is a
    product of functions.
    (b) y = e^2x^2
    Hint: Use the chain–rule with u = 2x^2.


    I have no clue how to solve these.help please is there a special equotation I have to use????

    Last edited by orsonik; November 16th 2007 at 02:03 PM. Reason: wrong font
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by orsonik View Post

    (a) y = e^x ln x + x^3
    Hint: Note that the first term on the right hand side is a
    product of functions.
    (b) y = e^2x^2
    Hint: Use the chain–rule with u = 2x^2.


    I have no clue how to solve these.help please is there a special equotation I have to use????

    First note that
    \frac{d}{dx}e^x = e^x

    \frac{d}{dx}ln(x) = \frac{1}{x}

    So
    y = e^x~ln(x) + x^3

    \frac{dy}{dx} = \left ( \frac{d}{dx} e^x \right ) ln(x) + e^x \frac{d}{dx}ln(x) + \frac{d}{dx}x^3

    \frac{dy}{dx} = e^x~ln(x) + e^x \cdot \frac{1}{x} + 3x^2

    For the second one:
    y = e^{2x^2}

    Let f(u) = e^u and u(x) = 2x^2.

    So
    y = f(u(x))

    \frac{dy}{dx} = \frac{df}{du} \cdot \frac{du}{dx}

    \frac{dy}{dx} = e^{u(x)} \cdot 4x

    \frac{dy}{dx} = 4x~e^{2x^2}

    -Dan
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  3. #3
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    yyyy

    oh my days. you are a real matematician magitian. I wish I was as good but I will probably never understand maths((
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