Bsrhrki

Yesterday a student in my calculus class attempted something like this:

Task: Derive $3^5x+1$ with respect to $x$.

Proposed solution: Let the inner function be $g(x)=3$ and the outer function be $f(z)=z^5x+1$, so that $f(g(x))=3^5x+1$. Then $f'(z)=(5x+1)cdot z^5x$ and $g'(x)=0$, so by the chain rule
$$
fracd (3^5x+1)dx = f'(g(x))g'(x)=0.
$$

I had difficulties to explain what's wrong with this and basically just told the student "the" right way to do it. Although I now have a rough idea of what's wrong, I'd like to hear from others:

1. Have you seen similar attempts?


2. How would you explain to a beginning calculus student what's wrong with this specific solution?

f is not a function of (only) z - f here is a function of x as well as z. I think this explanation is intelligible to a calc 1 student, and gets at the heart of the matter.

This is a VERY VERY typical problem. In fact, it's a problem even for $fracddx3^x$, much less your example.

The way I try to deal with this is one of two ways.

1. What has to happen first? To evaluate $3^5x+1$, you have to evaluate $5x+1$ first. So that is the inside function in the chain rule, just like in $sin(x^2)$ you have $x^2$ to evaluate first, so it is the inside function.




2. You could rethink how we notate or talk about exponential functions. In particular, Excel has $e^x$ written as exp(x) (I think as an option). So one can ask what the "input" is here.

However, on the plus side the student does seem to have the chain rule down; it's just the exponential notation that is causing trouble. So there is definitely hope here. And again, you should not be surprised at encountering this, so it is worth your time to come up with several possible responses for it in the long run. Good luck!

The root of the difficulty is that $x$ appears free in $f(z)$, but we are trying to "capture" it with $g(x)$, which is illegal. When we substitute $g(x)$ into $f(g(x))$, we have a variable clash:
$$
f(g(colorred x)) = 3^5colorbluex + 1
$$

The red (first) $x$ is a different variable from the blue (second) $x$. This is clearer if we rename the bound variable:
$$
f(g(colorred y)) = 3^5colorbluex + 1
$$

The original expression had $x$ bound to the $mathrm d x$, so by unbinding it, we have changed the meaning of the expression:
$$
fracmathrm dmathrm d colorblue x f(g(colorredy)) ne fracmathrm dmathrm d colorredy f(g(colorredy))
$$

(Incidentally, this is one reason I dislike the notation $f'(x)$, because it hides the variable of differentiation. Students must still be taught it, unfortunately, because Leibniz's notation is verbose in some contexts, but it should only be used as shorthand. Students should understand that it is a shorthand, and that there is still a variable of differentiation, even if it is not shown.)