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Math Help - derivative help

  1. #1
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    derivative help

    I'm having trouble figuring out how to find the derivative of xcosx. I know the answer is xcosx (-sinx lnx + (cosx)/x) but i have no idea how to get there.
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  2. #2
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    Re: derivative help

    Quote Originally Posted by JML2618 View Post
    I'm having trouble figuring out how to find the derivative of xcosx. I know the answer is xcosx (-sinx lnx + (cosx)/x) but i have no idea how to get there.
    x^{y}=e^{ln[x]y}

    see if that gets you started
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  3. #3
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    Re: derivative help

    Quote Originally Posted by JML2618 View Post
    I'm having trouble figuring out how to find the derivative of xcosx. I know the answer is xcosx (-sinx lnx + (cosx)/x) but i have no idea how to get there.
    I also encouraged student to study the following and be able to tell me by way of logarithmic differentiation why it is true.

    Suppose that each of f~\&~g is a differential function, then
    {D_x}\left( {g{{(x)}^{f(x)}}} \right) = \left( {g{{(x)}^{f(x)}}} \right)\left( {f'(x)\log (g(x)) + f(x)\frac{{g'(x)}}{{g(x)}}} \right)
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  4. #4
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    Re: derivative help

    I would have done it as "if y= x^{cos(x)} then ln(y)= cos(x) ln(x) so that \frac{1}{y}y'= -sin(x)ln(x)+ \frac{cos(x)}{x} and then
    y'= y\left(-sin(x)ln(x)+ \frac{cos(x)}{x}\right)= x^{cos(x)}\left(-sin(x)ln(x)+ \frac{cos(x)}{x}\right)".
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  5. #5
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    Re: derivative help

    Thanks for the replies they helped and i understand it now!
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