# Thread: A weird situation in calculus (product rule)?

1. ## 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.

2. ## Re: 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.
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*}

3. ## 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!

4. ## 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!
From scratch - it's a method called "Separation of Variables", basically it means getting each side to be only in terms of one variable.

5. ## 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?

6. ## 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?
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

7. ## 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)

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

Originally Posted by biffboy
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