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.

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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.