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Math Help - Apply the generalized product rule.

  1. #1
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    Apply the generalized product rule.

    x^x

    Online Derivative Calculator ? Shows All Steps!

    My derivative calculator says to apply the generalized product rule so that:

    x^x*\frac{d}{dx}(ln(x)*x)

    At first I thought it looked kind of like the chain rule however x^x is not the derivative of x^x. The answer output by the derivative calculator matches my solutions manual but I don't understand the steps I think.

    Thanks for any responses...

    P.S. I the calculator just means the product rule then I am still confused about ln(x) and how ln(x) became a part of the equation.
    Last edited by sepoto; November 26th 2013 at 08:14 PM.
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  2. #2
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    Re: Apply the generalized product rule.

    Quote Originally Posted by sepoto View Post
    x^x

    Online Derivative Calculator ? Shows All Steps!

    My derivative calculator says to apply the generalized product rule so that:

    x^x*\frac{d}{dx}(ln(x)*x)

    At first I thought it looked kind of like the chain rule however x^x is not the derivative of x^x. The answer output by the derivative calculator matches my solutions manual but I don't understand the steps I think.

    Thanks for any responses...

    P.S. I the calculator just means the product rule then I am still confused about ln(x) and how ln(x) became a part of the equation.
    note that xx = ex ln(x)

    and that d/dx (ef(x)) = ef(x)*df/dx(x)

    that should be all you need to compute your desired derivative.
    Thanks from topsquark and sepoto
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    Re: Apply the generalized product rule.

    e^{xln(x)}

    I am wondering how the equation above became equivalent to [tex]x^x[tex]. Also which x in the equation with e came from the exponent in the original x^x?
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    Re: Apply the generalized product rule.

    Quote Originally Posted by sepoto View Post
    e^{xln(x)}

    I am wondering how the equation above became equivalent to [tex]x^x[tex]. Also which x in the equation with e came from the exponent in the original x^x?
    e^{x \cdot ln(x)} = \left ( e^{ln(x)} \right ) ^x = x^x

    -Dan
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    Re: Apply the generalized product rule.

    Thank you for clearing that up. So then:

    (e^{xln(x)})'=(xlnx)'e^{xln(x)}????

    That's interesting so I think the above looks a little bit like the chain rule but I'm not sure yet about that.
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    Re: Apply the generalized product rule.

    It is the chain rule. letting y= x ln(x)= x^x, e^{xln(x)}= e^y.

    \frac{de^y}{dx}= \frac{de^y}{dy}\frac{dy}{dx}= \frac{dy}{dx}e^y

    dy/dx= (x ln(x))' and e^y= e^{x ln(x)} so (e^{x ln(x)})'= (x ln(x))' e^{x ln(x)}
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    Re: Apply the generalized product rule.

    So the derivative of e^x is e^x

    I am seeing the chain rule as:

    (e^{x*ln(x)})'*(x*ln(x))'

    Is that above not correct?

    I am trying to see how that might be similar to:

    (x*ln(x))'e^{x*ln(x)}

    The:
    e^{x*ln(x)}

    is confusing because I don't see that above as a derivative.


    Thanks for any responses...
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  8. #8
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    Re: Apply the generalized product rule.

    Quote Originally Posted by sepoto View Post
    So the derivative of e^x is e^x

    I am seeing the chain rule as:

    (e^{x*ln(x)})'*(x*ln(x))'

    Is that above not correct?
    No, this isn't correct

    (ef(x))' = f'(x) ef(x)
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    Re: Apply the generalized product rule.

    Thank you for clearing that up. I believe that is true but I am still confused about something. From what I know about the chain rule it is the derivative of the whole part multiplied by the derivative of the inside.

    Example:
    https://www.math.ucdavis.edu/~kouba/....html#SOLUTION1

    ((3x+1)^2)'*(3x+1)'=f'

    In the case f'(x)*e^{f(x)} my question is that e^{f(x)} does not appear to be a derivative in this case. So f'(x) clearly is a derivative but e^{f(x)} does not look like one. Is there something I don't understand maybe about the chain rule?

    Thanks for any responses....
    Last edited by sepoto; November 27th 2013 at 01:46 PM.
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    Re: Apply the generalized product rule.

    Quote Originally Posted by sepoto View Post
    Thank you for clearing that up. I believe that is true but I am still confused about something. From what I know about the chain rule it is the derivative of the whole part multiplied by the derivative of the inside.

    Example:
    https://www.math.ucdavis.edu/~kouba/....html#SOLUTION1

    ((3x+1)^2)'*(3x+1)'=f'

    In the case f'(x)*e^{f(x)} my question is that e^{f(x)} does not appear to be a derivative in this case. So f'(x) clearly is a derivative but e^{f(x)} does not look like one. Is there something I don't understand maybe about the chain rule?

    Thanks for any responses....
    it might be clearer if you thought of the example about as follows.

    d/dx((3x+1)2) = d/dx(u2) where u = 3x+1

    d/dx(u2) = d/du(u2) du/dx = 2u * 3 = 2(3x+1)*3 = 6(3x+1)

    Apply this thinking to your exponential example

    d/dx (ef(x))

    let u = f(x) du/dx = f'(x)

    d/dx(eu) = d/du(eu) du/dx = eu f'(x) = ef(x) f'(x)
    Thanks from sepoto
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