## Cute Proofs: The Product Rule for Derivatives.

### November 3, 2010

$\mbox{\tiny Two functions (solid blue and red) make the dotted purple function when multiplied!}$

There’s always a few calculus students who make the error of trying to take the derivative of $fg$ and get $f'g'$.  Of course, we know that this is not true in general, and the product rule for derivatives is as follows:

Theorem (Product Rule).  If $f, g$ are differentiable, then $(fg)' = f'g + g'f$

Given this formula, it’s a nice exercise for students to find out for which functions it is true that $(fg)' = f'g'$

Anyway, the proof of this theorem is not too difficult, even for calculus students, so I write it out for anyone interested.  This proof is really a "follow your nose" one with only one tricky part.  Let’s do it!

Proof.  We want to find $(fg)'$, so let’s set up the normal difference quotient.

$(fg)'(x) = \lim_{h\rightarrow 0}\frac{f(x + h)g(x+h) - f(x)g(x)}{h}$

Now the tricky part: we add and subtract $f(x)g(x)$ on the top.  This is really adding 0, since we’re going to add and subtract it, so this does not change our limit at all.  Thus,

$(fg)'(x) = \lim_{h\rightarrow 0}\frac{f(x + h)g(x+h) +f(x)g(x) - f(x)g(x)- f(x)g(x)}{h}$

$= \lim_{h\rightarrow 0}\frac{f(x)[g(x+h) - g(x)] + g(x+h)[f(x+h)- f(x)]}{h}$

$= \left[ \lim_{h\rightarrow 0} f(x)\right]\lim_{h\rightarrow 0}\frac{[g(x+h) - g(x)]}{h}$

$+ \left[\lim_{h\rightarrow 0}g(x+h)\right]\frac{[f(x+h)- f(x)]}{h}$

We can split up these limits because we know the limit of $f$ exists and we know the derivative of $g$ exists, and so if the limit of two functions exist and they are multiplied, the product of their limits is the limit of their products.  Anyway, this gives us

$= f(x)g'(x) + f'(x)g(x)$

which is exactly what we wanted to show.  $\Box$

This proof is pretty cute, I’m not gonna lie, and the only two tough parts were the product of two limits part and the initial adding and subtracting a common term part.

## Aren’t you Bored to Death of that Proof, though?

While looking for a decent proof of this, I stumbled upon a really sweet exercise that said, "Use logarithms to prove a version of the product rule."  I’d never seen this done before, so I thought I’d share it.  In this case, we can only have that our functions are positive, which sort of restricts us a bit, but we note that since pretty much every function we care about in calculus can be written as the difference of two positive functions, and since $f,g$ have to be differentiable and therefore continuous for the product rule to work in general, this isn’t so bad of a restriction.

Proof.  Suppose that $h(x) = f(x)g(x)$.  This is equivalent to saying

$\ln(h(x)) = \ln(f(x)g(x)) = \ln(f(x)) + \ln(g(x))$

by the property of logs.  Now, take the derivative of both sides.

$\dfrac{1}{h(x)}\dfrac{dh}{dx} = \dfrac{1}{f(x)}\dfrac{df}{dx} + \dfrac{1}{g(x)}\dfrac{dg}{dx}$

Now we multiply both sides by $h = fg$, and we obtain

$\dfrac{dh}{dx} = g(x)\dfrac{df}{dx} + f(x)\dfrac{dg}{dx}$

which is exactly the product rule.  $\Box$