# A better demonstration of the product rule

Inspired by The Theoretical Minimum: What You Need to Know to Start Doing Physics: here’s a better proof/justification for the product rule in differential calculus than the one I set out here last month.

We will start with what we will treat as an axiomatic definition of the differential of the function $y=f(x)$:

$\frac{dy}{dx} = \frac{df(x)}{dx} = \frac{f(x+\Delta x) - f(x)}{\Delta x}$ as $\Delta x \rightarrow 0$

In this case we have $y=f(x)g(x)$, so $\frac{dy}{dx} = \frac{f(x + \Delta x)g(x +\Delta x) - f(x)g(x)}{\Delta x}$

From our definition we can substitute for $f(x+\Delta x)$ and $g(x + \Delta x)$ and simplifying our notation for presentational reasons so that $\frac{df(x)}{dx} = f^{\prime}$ etc:

$f(x+\Delta x) = f^{\prime}\Delta x + f(x)$

$g(x+\Delta x) = g^{\prime}\Delta x + g(x)$

Giving (after dividing through by $\Delta x$):

$y^{\prime} =f^{\prime}g^{\prime}\Delta x + g(x)f^{\prime} + \frac{f(x)g(x)}{\Delta x} + g^{\prime}f(x) - \frac{f(x)g(x)}{\Delta x}$

$=f^{\prime}g^{\prime}\Delta x + g(x)f^{\prime} +g^{\prime}f(x)$

As $\Delta x \rightarrow 0$ the first term falls to zero and so we are left with:

$y^{\prime}=f^{\prime}g(x) + g^{\prime}f(x)$

Which, of course, is the product rule.

Update: See this most excellent comment from Professor Rubin.

# Another Euler number proof

OK, back to the issue of Euler’s number and the proof that it is $\lim _{x \to \infty}(1 + \frac{1}{x})^x$. This is a proof based on, though expanded from, what I picked up from a rather excellent website here.

Once again we start from the proposition that there exists a number $e$ such that $\frac{d}{dx}e^x = e^x$ and hence $\frac{d}{dx}log_e(x) = \frac{1}{x}$ (see here for why the second follows from the first).

Here we go…

$y = \lim_{x \to \infty}(1 + \frac{1}{x})^x$

$log_e(y) = log_e[\lim_{x \to \infty}(1 + \frac{1}{x})^x]$

$= \lim_{x \to \infty}[x log_e(1 +\frac{1}{x})]$

At this point we still have $\infty \times 0$, an indeterminate number, so we need to look for a determinate form.

So we take this as:

$= \lim_{x \to \infty}[\frac{log_e(1 + \frac{1}{x})}{\frac{1}{x}}]$

Now we have $\frac{0}{0}$, another indeterminate, but we can also apply L’Hospital’s rule.

This states that, $\lim_{x \to \pm\infty}[\frac{\frac{d}{dx}f(x)}{\frac{d}{dx}g(x)} = \frac{f(x)}{g(x)}]$ (a proof to follow sometime).

Here $f(x) = log_e(1+\frac{1}{x})$, $g(x) = \frac{1}{x}$ and, of course, $\lim_{x \to \infty}\frac{f(x)}{g(x)}= log_e[\lim_{x \to \infty}(1+\frac{1}{x})^x]$.

So, using the chain rule, $\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}$, for the numerator, $y = log_e(u)$, $u=1+\frac{1}{x}$: so $\frac{d}{dx}log_e(1 + \frac{1}{x}) = \frac{1}{1 + \frac{1}{x}}\times-\frac{1}{x^2}$

And for the denominator $\frac{d}{dx}\frac{1}{x} = -\frac{1}{x^2}$, giving for the whole thing, $\lim_{x \to \infty}[\frac{1}{1+\frac{1}{x}}] =1$

Hence $log_e[\lim_{x \to \infty}(1 + \frac{1}{x})^x] = 1$, so $e=\lim_{x \to \infty}(1 +\frac{1}{x})^x$.

# Differential calculus reminder

This is just an online note to myself about differential calculus. A level maths again…

Calculating $\frac{d}{dx}2^x$

$2^x = e^u$ where $u=\ln(2^x)$

Using the chain rule: $\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}$

$\frac{dy}{du} = e^u$ (as $\frac{d}{dx}e^x = e^x$)

$\ln(2^x) = x\ln(2)$ hence $\frac{d}{dx}\ln(2^x) = \ln(2)$.

So $\frac{d}{dx}2^x = e^{\ln(2^x)}\ln(2) = 2^x\ln(2)$

And $e$, Euler’s number, really is magical.