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 $\displaystyle \displaystyle \begin{align*} \frac{d}{dx}\left(f\,g\right) = \frac{df}{dx} \,\frac{dg}{dx} \end{align*}$ we have
$\displaystyle \displaystyle \begin{align*} \frac{d}{dx}\left(f\,g\right) &= \frac{df}{dx} \, \frac{dg}{dx} \\ f\,\frac{dg}{dx} + g\,\frac{df}{dx} &= \frac{df}{dx} \, \frac{dg}{dx} \\ g\,\frac{df}{dx} &= \frac{df}{dx}\,\frac{dg}{dx} - f\,\frac{dg}{dx} \\ g\,\frac{df}{dx} &= \frac{dg}{dx}\left(\frac{df}{dx} - f\right) \\ \frac{1}{\frac{df}{dx} - f}\,\frac{df}{dx} &= \frac{1}{g}\,\frac{dg}{dx} \\ \frac{\frac{df}{dx}}{\frac{df}{dx} - f} &= \frac{1}{g}\,\frac{dg}{dx} \\ \frac{\frac{df}{dx} - f + f}{\frac{df}{dx} - f} &= \frac{1}{g}\,\frac{dg}{dx} \\ 1 + \frac{f}{\frac{df}{dx} - f} &= \frac{1}{g}\,\frac{dg}{dx} \\ \int{\left(1 + \frac{f}{\frac{df}{dx} - f}\right)dx} &= \int{\frac{1}{g}\,\frac{dg}{dx}\,dx} \\ \int{\left(1 + \frac{f}{\frac{df}{dx} - f}\right)dx} &= \int{\frac{1}{g}\,dg} \end{align*}$
$\displaystyle \displaystyle \begin{align*} \int{\left(1 + \frac{f}{\frac{df}{dx} - f}\right)dx} &= \ln{|g|} + C \\ \ln{|g|} &= \int{\left(1 + \frac{f}{\frac{df}{dx} - f}\right)dx} - C \\ |g| &= e^{\int{\left(1 + \frac{f}{\frac{df}{dx} - f}\right)dx} - C} \\ |g| &= e^{-C}e^{\int{\left(1 + \frac{f}{\frac{df}{dx} - f}\right)dx}} \\ g &= A\,e^{\int{\left(1+\frac{f}{\frac{df}{dx}-f}\right)dx}} \textrm{ where } A = \pm e^{-C}\end{align*}$
There you go, any function $\displaystyle \displaystyle \begin{align*} f\,g \end{align*}$ that has the property of $\displaystyle \displaystyle \begin{align*} g = A\,e^{\int{\left(1+\frac{f}{\frac{df}{dx}-f}\right)dx}} \end{align*}$ will result in $\displaystyle \displaystyle \begin{align*} \frac{d}{dx}\left(f\,g\right) = \frac{df}{dx}\,\frac{dg}{dx} \end{align*}$