College Math Teaching

October 4, 2018

When is it ok to lie to students? part I

Filed under: calculus, derivatives, pedagogy — collegemathteaching @ 9:32 pm

We’ve arrived at logarithms in our calculus class, and, of course, I explained that ln(ab) = ln(a) + ln(b) only holds for a, b > 0 . That is all well and good.
And yes, I explained that expressions like f(x)^{g(x)} only makes sense when f(x) > 0

But then I went ahead and did a problem of the following type: given f(x) = \frac{x^3 e^{x^2} cos(x)}{x^4 + 1} by using logarithmic differentiation,

f'(x) = \frac{x^3 e^{x^2} cos(x)}{x^4 + 1} (\frac{3}{x} + 2x -tan(x) -\frac{4x^3}{x^4+ 1})

And you KNOW exactly what I did. Right?

Note that f is differentiable for all x and, well, the derivative *should* be continuous for all x but..is it? Well, up to inessential singularities, it is. You see: the second factor is not defined for x = 0, x = \frac{\pi}{2} \pm k \pi , etc.

Well, let’s multiply it out and obtain:
f'(x) = \frac{3x^2 e^{x^2} cos(x)}{x^4 + 1} + \frac{2x^4 e^{x^2} cos(x)}{x^4 + 1} - \frac{x^3 e^{x^2} sin(x)}{x^4 + 1}-\frac{4x^6 e^{x^2} cos(x)}{(x^4 + 1)^2}

So, there is that. We might induce inessential singularities.

And there is the following: in the process of finding the derivative to begin with we did:

ln(\frac{x^3 e^{x^2} cos(x)}{x^4 + 1}) = ln(x^3) + ln(e^{x^2}) + ln(cos(x)) - ln(x^4 + 1) and that expansion is valid only for
x \in (0, \frac{\pi}{2}) \cup (\frac{5\pi}{2}, \frac{7\pi}{2}) \cup .... because we need x^3 > 0 and cos(x) > 0 .

But the derivative formula works anyway. So what is the formula?

It is: if f = \prod_{j=1}^k f_j where f_j is differentiable, then f' = \sum_{i=1}^k f'_i \prod_{j =1, j \neq i}^k f_j and verifying this is an easy exercise in induction.

But the logarithmic differentiation is really just a motivating idea that works for positive functions.

To make this complete: we’ll now tackle y = f(x)^{g(x)} where it is essential that f(x) > 0 .

Rewrite y = e^{ln(f(x)^{g(x)})} = e^{g(x)ln(f(x))}

Then y' = e^{g(x)ln(f(x))} (g'(x) ln(f(x)) + g(x) \frac{f'(x)}{f(x)}) = f(x)^{g(x)}(g'(x) ln(f(x)) + g(x) \frac{f'(x)}{f(x)})

This formula is a bit of a universal one. Let’s examine two special cases.

Suppose g(x) = k some constant. Then g'(x) =0 and the formula becomes y = f(x)^k(k \frac{f'(x)}{f(x)}) = kf(x)^{k-1}f'(x) which is just the usual constant power rule with the chain rule.

Now suppose f(x) = a for some positive constant. Then f'(x) = 0 and the formula becomes y = a^{g(x)}(ln(a)g'(x)) which is the usual exponential function differentiation formula combined with the chain rule.

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