# 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.