$\frac{d}{dx}(f(x)g(x)) \neq f^\prime(x)g^\prime(x)$? Think again!

One of the most unintuitive and frustrating facts about calculus is that \[\frac{d}{dx}(f(x)g(x))\neq f^\prime(x)g^\prime(x).\]
It is said in certain apocryphal tales that the great man himself once made the above mistake. Instead of the pretty but untrue equation above, we have the unsightly but true "product rule":
\[\frac{d}{dx}f(x)g(x)= f(x)g^\prime(x)+f^\prime(x)g(x).\]

If you're like me, you'll despise this terrible rule for all the extra algebra it usually introduces into typical calculus questions. My wrists would often hurt from applying and reapplying the product rule in repeated differentiations. So, with a bit of free time and a a desire for vengeance, I have compiled a list of functions (zoom in if you cannot see some of the variables) for which the equation \[\frac{d}{dx}(f(x)g(x))=f^\prime(x)g^\prime(x)\] is absolutely positively true. Take that product rule (all c's, n's, and a's are arbitrary constants)!

• \[f(x)=c_{1}, g(x)=c_{2}\]
• \[f(x)=c_{1}(a+nx)^{\frac{1}{n}}, g(x)=c_{2}(a+nx-1)^{-\frac{1}{n}}\]
• \[f(x)=c_{1}e^{nx}, g(x)=c_{2}e^{\frac{n}{n-1}x}\]
• \[f(x)=c_{1}e^{x}(a^{nx}-1)^{-\frac{1}{n\log(a)}}, g(x)=c_{2}e^{-\frac{a^{-nx}}{n\log(a)}}\]

"Pfft," you say. "I could have guessed those ones!" Well, let's make things a little more exciting and introduce some trig.

• \[f(x)=c_{1}\left(1-a\arccos(\sin(nx))\right)^{\frac{\sqrt{\cos^{2}(nx)}\sec(nx)}{an}}, g(x)=c_{1}\arccos(\sin(nx))^{-\frac{\sqrt{\cos^{2}(nx)}sin(nx)}{an}}\]
• \[f(x)=c_{1}e^{\frac{\coth(a+nx)}{n}}, g(x)=c_{2}e^{\frac{\tanh(a+nx)}{n}}\]
• \[f(x)=c_{1}e^{\frac{-\cot(a+nx)}{n}}, g(x)=c_{2}e^{\frac{\tan(a+nx)}{n}}\]

That wasn't much. Let's step it a notch (at this point you will appreciate the ability on this blog to zoom on the formulae to any degree).

• \[f(x)=c_{1}e^{\frac{2\arctan\left(\tanh\left(\frac{a+nx}{2}\right)\right)}{n}}, g(x)=c_{1}e^{\frac{\coth\left(\frac{a+nx}{2}\right)}{n}}\]
• \[f(x)=c_{1}e^{\frac{2(a+nx)+\sin(2(a+nx))}{4n}}, g(x)=c_{2}e^{\frac{\cot(a+nx)}{n}+x}\]

Finally, I found these gems after a bit of searching. Beautiful, right?

• \[f(x)=c_{1}\exp\left(-\frac{2\arctan\left(\frac{a-\tan\left(\frac{nx}{2}\right)}{\sqrt{1-a^{2}}}\right)}{n\,\sqrt{1-a^{2}}}\right), g(x)=c_{2}\sin\left(\frac{nx}{2}\right)^{\frac{1}{an}}\cos\left(\frac{nx}{2}\right)^{-\frac{1}{an}}\]
• \[f(x)=c_{1}\exp\left(\frac{2\sin\left(\frac{a+nx}{2}\right)\left(\csc\left(\frac{a}{2}\right)\sin\left(\frac{nx}{2}\right)-nx\sin\left(\frac{a+nx}{2}\right)\right)}{n(\cos(a+nx)-1)}\right), g(x)=c_{2}e^{\frac{\sin(a+nx)}{n}}\]
• \[f(x)=c_{1}\exp\left(\frac{nx-a\log(a\sin(nx)-\cos(nx))}{a^{2}n+n}\right), g(x)=c_{2}\sin(nx)^{\frac{1}{an}}\]

These functions all obey the "nice" product rule \[\frac{d}{dx}(f(x)g(x))=f^\prime(x)g^\prime(x)\] with typical restrictions such not allowing constants which cause divisions by zero.

So how did I do it?
A bit of calculus and creativity.

Using the product rule on the original equation:
\[(f(x)g(x))^{\prime}=f^{\prime}(x)g^{\prime}(x)\] \[f(x)g^{\prime}(x)+f^{\prime}(x)g(x)=f^{\prime}(x)g^{\prime}(x)\]

Dividing both sides by $f^{\prime}(x)g^{\prime}(x)$ we obtain:

\[\frac{f(x)}{f^{\prime}(x)}+\frac{g(x)}{g^{\prime}(x)}=1\]
Now comes the creative part. I pick a function, say $x+(1-x)=1$. Let $\frac{f(x)}{f^{\prime}(x)}=x$, $\frac{g(x)}{g^{\prime}(x)}=1-x$

We now have two differential equations we can solve:

\[\frac{f}{f^\prime}=x,\frac{g}{g^\prime}=1-x\]

Solving these differential equations we obtain:

\[f(x)=c_{1}x,g(x)=\frac{c_{2}}{1-x}\]

which obey the nice equation \[\frac{d}{dx}(f(x)g(x))= f^\prime(x)g^\prime(x).\]

The more functions that you can think of which adds up 1, the more functions you can produce which obey our "nice" product rule.

I'd love to see what you guys can come up with, so post your functions in the comments below! To make it more challenging, try to make the functions you find as general as possible, and only find functions which are composed of elementary functions!