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Math Help - A weird situation in calculus (product rule)?

  1. #1
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    A weird situation in calculus (product rule)?

    So how many times have you tried doing (fg)' and instead of doing fg' + f'g you directly did f'g'?

    Im just wondering if this situation ever really exists?

    For what functions is (fg)' = f'g'

    Thanks.
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  2. #2
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    Re: A weird situation in calculus (product rule)?

    Quote Originally Posted by federernadal View Post
    So how many times have you tried doing (fg)' and instead of doing fg' + f'g you directly did f'g'?

    Im just wondering if this situation ever really exists?

    For what functions is (fg)' = f'g'

    Thanks.
    Well let's see... If we let \displaystyle \begin{align*} \frac{d}{dx}\left(f\,g\right) = \frac{df}{dx} \,\frac{dg}{dx} \end{align*} we have

    \displaystyle \begin{align*} \frac{d}{dx}\left(f\,g\right) &= \frac{df}{dx} \, \frac{dg}{dx} \\ f\,\frac{dg}{dx} + g\,\frac{df}{dx} &= \frac{df}{dx} \, \frac{dg}{dx} \\ g\,\frac{df}{dx} &= \frac{df}{dx}\,\frac{dg}{dx} - f\,\frac{dg}{dx} \\ g\,\frac{df}{dx} &= \frac{dg}{dx}\left(\frac{df}{dx} - f\right) \\ \frac{1}{\frac{df}{dx} - f}\,\frac{df}{dx} &= \frac{1}{g}\,\frac{dg}{dx} \\ \frac{\frac{df}{dx}}{\frac{df}{dx} - f} &= \frac{1}{g}\,\frac{dg}{dx} \\ \frac{\frac{df}{dx} - f + f}{\frac{df}{dx} - f} &= \frac{1}{g}\,\frac{dg}{dx} \\ 1 + \frac{f}{\frac{df}{dx} - f} &= \frac{1}{g}\,\frac{dg}{dx} \\ \int{\left(1 + \frac{f}{\frac{df}{dx} - f}\right)dx} &= \int{\frac{1}{g}\,\frac{dg}{dx}\,dx} \\ \int{\left(1 + \frac{f}{\frac{df}{dx} - f}\right)dx} &= \int{\frac{1}{g}\,dg} \end{align*}

    \displaystyle \begin{align*} \int{\left(1 + \frac{f}{\frac{df}{dx} - f}\right)dx} &= \ln{|g|} + C \\ \ln{|g|} &= \int{\left(1 + \frac{f}{\frac{df}{dx} - f}\right)dx} - C \\ |g| &= e^{\int{\left(1 + \frac{f}{\frac{df}{dx} - f}\right)dx} - C} \\ |g| &= e^{-C}e^{\int{\left(1 + \frac{f}{\frac{df}{dx} - f}\right)dx}} \\ g &= A\,e^{\int{\left(1+\frac{f}{\frac{df}{dx}-f}\right)dx}} \textrm{ where } A = \pm e^{-C}\end{align*}

    There you go, any function \displaystyle \begin{align*} f\,g \end{align*} that has the property of \displaystyle \begin{align*} g = A\,e^{\int{\left(1+\frac{f}{\frac{df}{dx}-f}\right)dx}} \end{align*} will result in \displaystyle \begin{align*} \frac{d}{dx}\left(f\,g\right) = \frac{df}{dx}\,\frac{dg}{dx} \end{align*}
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  3. #3
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    Re: A weird situation in calculus (product rule)?

    This is really impressive!!

    Did you do that from scratch or have you been asked this before?

    Thanks!
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    Re: A weird situation in calculus (product rule)?

    Quote Originally Posted by federernadal View Post
    This is really impressive!!

    Did you do that from scratch or have you been asked this before?

    Thanks!
    From scratch - it's a method called "Separation of Variables", basically it means getting each side to be only in terms of one variable.
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  5. #5
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    Re: A weird situation in calculus (product rule)?

    Yup I'm in Diff Eq class now. I had no trouble following your method, but it would be difficult for me to come up with it from scratch.

    You are a professor at which university?
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    Re: A weird situation in calculus (product rule)?

    Quote Originally Posted by federernadal View Post
    Yup I'm in Diff Eq class now. I had no trouble following your method, but it would be difficult for me to come up with it from scratch.

    You are a professor at which university?
    I do teach at a university (RMIT University in Melbourne, Australia) but I'm far from being a Professor. I'm still working on my Masters :P
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    Re: A weird situation in calculus (product rule)?

    Great stuff. Not seen that done before. Looks really neat if we simplify bracket to f'/(f'-f)
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    Re: A weird situation in calculus (product rule)?

    Quote Originally Posted by biffboy View Post
    Great stuff. Not seen that done before. Looks really neat if we simplify bracket to f'/(f'-f)
    I prefer to only have one derivative in it, which is why I wrote it as I did
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