blog.md

title	author	date	output
Why is Implicit Differentation So Hard?	Zach Fabry	12/22/2020	pdf_document

Implicit differentiation--the scourge of first-time calculus students everywhere. Even if you manage to master derivatives as you are taught them, implicit differentiation can be confusing. Many people still don't have a good intuition as to how it works by the end of their first calculus course (myself included), even if they learn to perform the operation successfully. Why is this? Why does it seem so hard?

A quick review of implicit differentiation

Say we are given the equation $y = x^2 + 3x$ and asked to differentiate it with respect to $x$. This is the form of differentiation problem we are all likely most familiar with; though we normally just refer to it as differentiation, it is technically \textit{explicit} differentiation, since $y$ is solved for explicitly (i.e. the equation is in $y = ;$ form). A straightforward application of derivative rules tells us that the derivative in this case is $$\frac{dy}{dx} = 2x + 3 ; .$$ However, there are times when we either cannot solve for $y$ explicitly or it would be time-consuming to do so; this is where implicit differentiation can be useful. Suppose that instead we are asked to differentiate the equation $x^2 + y^2 = 36$, so that $y$ is no longer explicitly defined. Differentiating with respect to $x$ again $$\frac{d}{dx} \left[ x^2 + y^2 = 36 \right] $$ yields $$2x + 2y \frac{dy}{dx} = 0 $$ which gives us $$ \frac{dy}{dx} = - \frac{x}{y}$$ after rearranging to solve for $\frac{dy}{dx}$ explicitly. Why the $\frac{dy}{dx}$ term appears in the first place, though, can be unintuitive; it's usually briefly mentioned as a consequence of the chain rule, and then swiftly forgotten about.

The chain rule

The chain rule allows us to differentiate compositions of functions, i.e. functions of the form $f(g(x))$. It's probably most often seen in that function/prime notation as $$\frac{d}{dx} \left[ ; f(g(x)) ; \right] = f'(g(x)) \cdot g'(x)$$ but in the Leibniz notation we have been using up until now it is written as $$\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$$ for $y = f(u)$ and $u = g(x)$. Revisiting our implicit differentiation problem of $$\frac{d}{dx} \left[ x^2 + y^2 = 36 \right] $$ and assuming $y$ is a differentiable function of $x$ (which is always assumed for implicit differentiation problems, but rarely mentioned), we then use the chain rule to see that $$\frac{d}{dx} \left[ y^2 \right] = \frac{d}{dy} \left[ y^2 \right] \frac{dy}{dx} = 2y \frac{dy}{dx} ; ,$$ which explains where that $\frac{dy}{dx}$ term comes from. This has always felt vaguely unsatisfying to me, though, and I know I'm not alone in having struggled to wrap my head around it the first few times (and several more after that).

A better way?

I have found success thinking about this and helping others to think about it with what I believe is a relatively simple solution--we should simply use the chain rule all the time, \textit{even before we know it as the chain rule}. We accept without much questioning that $\frac{d}{dx} [y] = \frac{dy}{dx}$; by the same logic, it should follow that $\frac{d}{dx} [x] = \frac{dx}{dx}$, but this is rarely written out. If we think of it as a ratio (whether or not it technically is a ratio is beyond the point here), this tracks with what we know--that $\frac{d}{dx} [x] = 1$. If we don't want to think of it as a ratio, or for powers of $x$ greater than 1, we could simply think of it as keeping track of what variable we are differentiating with respect to; this might be less technically correct, but exposure to the notation earlier might make implicit differentiation (and the chain rule in general) less intimidating. Working through our earlier explicit differentiation example again with this notation, we now have $$\frac{d}{dx} \left[ y = x^2 + 3x \right]$$ give us $$\frac{dy}{dx} = 2x \frac{dx}{dx} + 3 \frac{dx}{dx} = 2x(1) + 3(1) = 2x + 3 ; ,$$ which is of course identical to earlier, at the cost of just a few extra symbols written. Similarly, our implicit differentiation solution now becomes $$\frac{d}{dx} \left[ x^2 + y^2 = 36 \right] $$ which gives $$2x \frac{dx}{dx} + 2y \frac{dy}{dx} = 0 \frac{dx}{dx} $$ which is $$2x(1) + 2y \frac{dy}{dx} = 0(1) $$ and we are again back to where we earlier-- $$2x + 2y \frac{dy}{dx} = 0 $$ and the desired solution of $$ \frac{dy}{dx} = - \frac{x}{y}$$ after rearranging.

This certainly isn't a new idea, but I don't see it nearly often enough. Most calculus books I have seen make no mention of using $\frac{dx}{dx}$; occasionally it is mentioned when doing implicit differentiation, but almost never when first discussing derivatives. This seems like a mistake to me; I have had a fair amount of success describing it this way to students who are struggling with implicit differentiation, and multiple times people have told me that they wish it was taught this way in the beginning, as it would have saved them some headache when implicit differentiation came up. That alone makes me think it should at least be mentioned, even if people decide not to write it all out every single time. It can help with related rates problems, too; a derivative of the form $$\frac{d}{dt} \left[ V = \frac{4}{3} \pi r^3 \right]$$ giving $$\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$$ seems a lot less mysterious if we think about just tracking what variable we are differentiating with respect to (if we want to use notation tricks to better complete the process), or if we are more familiar with the chain rule from the beginning (if we want to be more technically correct). Either way, it seems like a missed opportunity to help students understand things more easily that this is not in more calculus books and classes.