How to explain what's wrong with this application of the chain rule?

Question

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

Problem statement: Find the derivative of $3^{(5x+1)}$ with respect to $x$.

Proposed solution:

• Let the inner function be given by $g(x)=3,$ and the outer by $f(z)=z^{(5x+1)}$, so that $$f(g(x))=3^{(5x+1)}.$$
• $f'(z)=(5x+1)\cdot z^{(5x)}$ and $g'(x)=0$, so by the chain rule, $$\frac{d\left(3^{(5x+1)}\right)}{dx} = f'(g(x))g'(x)=0.$$

I had difficulties explaining 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?

Answer 1

$$ \frac{d (3^{5x+1})}{dx} = f'(g(x))g'(x)= \frac{d \left(3^{5x+1}\right)}{d(3)} \times \frac{d (3)}{dx}. $$

However $\dfrac{d (3^{5x+1})}{d(3)}$ is undefined.

Answer 2

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: $$ f(g(\color{red} x)) = 3^{5\color{blue}x + 1} $$

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

The original expression had $x$ bound to the $\mathrm d x$, so by unbinding it, we have changed the meaning of the expression: $$ \frac{\mathrm d}{\mathrm d \color{blue} x} f(g(\color{red}y)) \ne \frac{\mathrm d}{\mathrm d \color{red}y} f(g(\color{red}y)) $$

(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.)

Answer 3

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.

Answer 4

This is a VERY VERY typical problem. In fact, it's a problem even for $\frac{d}{dx}3^x$, 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!

Answer 5

The other answers have completely missed the mistake. $ \def\rr{\mathbb{R}} $

Your student's error has nothing to do with exponentiatiation. Consider the following based on exactly the same error:

$\color{red}{\text{Let (???)}}$ $f(y) = x$ and $g(x) = 1$.

Then $1 = \frac{dx}{dx} = (f∘g)'(x) = f'(g(x))·g'(x) = f'(1)·0 = 0$.

The error lies in the very first line! It is extremely obvious once you actually attempt to make it rigorous. Recall that to define a function you must provide a domain as well as a rule that specifies the output for each input in the domain. And of course the rule has to be meaningful in the context where you want to define the function. So see what you get:

$\color{red}{\text{Let (???)}}$ $f : \rr→\rr$ such that $f(y) = x$ for each $y∈\rr$.

Let $g : \rr→\rr$ such that $g(x) = 1$ for every $x∈\rr$.

The definition of $g$ is fine. The definition of $f$ is not fine! What on earth is $x$? The rule has to specify the output for each input $y∈\rr$, so where did $x$ pop up from?

As explained above, the error has nothing to do with differentiation. Rather, it is in the illegal definition of the function!

Answer 6

This idea is fine, and you can use the multivariable chain rule to do it this way.

Say we want to differentiate $h(x) = f(x)^{g(x)}$ with respect to $x$. Notice that we can write $h$ as the composite of $p: \mathbb{R} \to \mathbb{R}^2$ defined by $p(t) = (f(t),g(t))$ with the function $E: \mathbb{R}^2 \to \mathbb{R}$ defined by $E(u,v) = u^v$.

By the multivariable chain rule,

$$ \begin{align} Dh\big|_{x} &= DE\big|_{p(x)} \circ Dp\big|_{x}\\ &= \left.\begin{bmatrix} \frac{\partial E}{\partial u} & \frac{\partial E}{\partial v} \end{bmatrix} \right|_{(u,v) = (f(x),g(x))} \circ \left.\begin{bmatrix} \frac{\partial f}{\partial t} \\ \frac{\partial g}{\partial t}\end{bmatrix}\right|_{t = x}\\ &= \left.\begin{bmatrix} vu^{v-1} & \ln(u) u^v \end{bmatrix} \right|_{(u,v) = (f(x),g(x))} \circ \left.\begin{bmatrix} f'(t) \\ g'(t)\end{bmatrix}\right|_{t = x}\\ &= \begin{bmatrix} g(x)(f(x))^{g(x)-1} & \ln(f(x)) (f(x))^{g(x)} \end{bmatrix} \circ \begin{bmatrix} f'(x) \\ g'(x)\end{bmatrix}\\ &=f'(x)g(x)(f(x))^{g(x)-1}+g'(x)\ln(x)f(x)^{g(x)} \end{align} $$

Applying this to the problem in question, we see that $f'(x) =0$, so the first term disappears.

So, in a sense, the student was trying to apply the multivariable chain rule (using the two variables $z$ and $x$), but didn't know how to do that yet. So you could tell them it is a good approach, but they will learn how to properly execute that approach in calc 3.

Answer 7

I had a similar problem with a student last week and could not succinctly explain why she could not select $e$ as the 'inner' function here: $$f(x) = e^{8x +4}$$

The best explanation I have seen thus far, Paul's Notes, explains it in this way:

Recall that the 'outside' function is the last operation that we would perform in an evaluation. In this case if we were to evaluate this function the last operation would be the exponential. Therefore, the outside function is the exponential function and the inside function is its exponent.

Answer 8

Some users have expressed doubt at the validity of the accepted answer, so let me make it rigorous. To do that, we first need to make sense of the notation $\frac{dy}{dx}$. (Which is not a trivial task.)

We start by changing perspective: instead of thinking of variables $x$ and $y$ as numbers, we think of them as smooth real valued functions on some manifold $M$. So $x\colon M\to \mathbb{R}$ and $y\colon M\to \mathbb{R}$. You might rightly ask: why would we do that and which manifold $M$ are you talking about? The answer to the first question is: because I'd like to talk about the differentials $dy$, $dx$ and say everyday stuff like "$y$ is a constant" or "$y$ is a function of $x$". All of that is impossible inside first order logic + ZFC if we simply interpret $x$ and $y$ as elements of $\mathbb{R}$. Concerning the second question: think of $M$ as the physical state space underlying the problem we are trying to model and $x$ and $y$ as observables. If that sounds too unfamiliar: it's similar to how people in probability theory assume an underlying space of outcomes $\Omega$, in order to talk about random variables. (Which are the things we really care about and historically came before $\Omega$, just like $dx,dy$ historically came before manifolds, but I'm drifting of.)

So, having fixed the background manifold $M$, whenever you hear me say variable, what I mean is a thing of type $M\to \mathbb{R}$.

Definition. Given two variables $x$ and $y$, call $y$ a function of $x$ if $dx$ is not zero almost everywhere and there exists a variable $q\colon M \to \mathbb{R}$ such that $$ dy = q \cdot dx. $$

Intuitively, the equation $ dy = q \cdot dx $ says that the change of $y$ is determined by the change of $x$, i.e. that $y$ depends on $x$.

It's not hard to show that $q$ is uniquely determined by $x$ and $y$, hence we decide to denote it with $\frac{dy}{dx}$ and call it the derivative of $y$ wrt. $x$. It was originally called the differential coefficient, cause that's what it is.

According to this definition, $3^{5x+1}$ is a function of $x$ (assuming $x$ is a true variabel, i.e. $dx\neq 0$), but it's also a function of $5x+1$. On the other hand, $3^{5x+1}$ is not a function of $3$ since $d3=0$ (what we call a constant). In particular $\frac{d3^{5x+1}}{d3}$ is undefined, as user21820 has been pointing out emphatically.

We can now state the chain rule in Leibniz form

Theorem. If $z$ is a function of $y$ and $y$ is a function of $x$, then $z$ is also a function $x$ and their differential coefficients satisfy $$ \frac{dz}{dx}=\frac{dz}{dy}\cdot \frac{dy}{dx} $$

The proof of this is trivial.

From this perspective, what the student in my question was trying to do, was to let $z=3^{5x+1}$ and $y=3$. But the theorem does not apply, since $3^{5x+1}$ is not a function $3$. The same point of view can be used in the example discussed here, where a student attempted to differentiate $x^x$ by taking $x$ as inner function. Although that's allowed it just leads to $$ \frac{dx^x}{dx}=\frac{dx^x}{dx}\cdot \frac{dx}{dx} $$ which is of not much use.

This is not to say that I don't appreciate the other answers (also users 21820). Taemyr's is just one of the three perspectives that haven been proposed. It might seem like it needs a lot of background to make it rigorous. But consider that mathematicians understood this stuff for at least 200 years without requiring manifolds to formalize it. And consider that the other approaches also require quite some background to make them rigorous (like quantifiers, variable bindings, the idea of dummy/bound variables etc. or derivatives of functions of multiple variables). Each has it advantages and disadvantages and none seems more right than the others, methinks.

Answer 9

Applying the naive approach of a non-mathematician, to me the expression $z^{(5x+1)}$ points to a bivariate function,

$$f(z,x) = z^{(5x+1)}$$

(because "I see two variables in here"), and with $g(x) = 3$ we have defined

$$f(g(x), x) = 3^{(5x+1)}$$

Then

$$\frac {df(g(x),x)}{dx} = \frac {\partial f(g(x),x)}{\partial g(x)}\cdot \frac {dg(x)}{dx} + \frac {\partial f(g(x),x)}{\partial x}\cdot \frac {dx}{dx}$$

$$=\frac {\partial f(g(x),x)}{\partial g(x)}\cdot 0 + \frac {\partial f(g(x),x)}{\partial x}\cdot 1 = \frac {\partial f(g(x),x)}{\partial x} $$

$$=\frac {\partial}{\partial x} \left(3^{(5x+1)}\right) $$

This appears to be correct, although not useful, since we ended up back in the beginning. Am I doing something wrong here?

Answer 10

This answer may not be a lot different from the other answers, but here is how I would phrase it:

When we use the chain rule to compute $\tfrac{d}{dx} f(g(x))$, the dependence on $x$ should only be through the input $g$. In your function, $f(z) = z^{5x+1}$, the function $f$ depends on $x$ not only through $z$, but also through the $5x+1$ in the exponent. It would be clearer to write it as $f(x,z)$, so this issue would be clear. When you use the chain rule, your $g$ should be big enough to include everything that depends on $x$ inside it.

(Optional second paragraph) If you need to compute $\tfrac{d}{dx} f(x,g(x))$, there is a formula for that: $$\frac{d}{dx} f(x,g(x)) = \frac{\partial f}{\partial x} (x,g(x)) +\frac{dg}{dx}(x) \cdot \frac{\partial f}{\partial y} (x,g(x)).$$

Answer 11

In writing $f(z)=z^{5x+1}$, the student is (incorrectly) taking $x$ to be a constant.

Now, given also $g(z)=3$ (for all $z \in \mathbb R$), we have $f(g(z))=f(3)=3^{5x+1}$ (where again $x$ is a constant).

Observe that given a small change in $a$, $f(g(a))$ doesn't change. That is, $(fg)^\prime(a)=0$.

And this is indeed what we get when we apply the Chain Rule:

Since $f^\prime(z)=(5x+1)z^{5x}$ and $g^\prime(z)=0$, we have $$(fg)^\prime(z)=f^\prime(g(z))g^\prime(z)=(5x+1)[g(z)]^{5x}\times0=0.$$