Why High School Kids Should Learn Partial Derivatives

25 Mar, 2024

I always feel that introducing differentiation as the gradient of some curve is both geometrically intuitive and algebraically constrained. Limiting calculus to one-variable functions generally gives students very wrong idea about differentiation.

A good idea to start differentiation is to think of the reaction to a small (infinitesimal) disturbance to input, that is

d f \approx α d x .

Naturally, for functions with more than one input (variable), we will assume that the cross-action is negligible when disturbance in each direction is small. Thus,

d f \approx \sum_{i} α_{i} x_{i} .

To obtain these $α_{i}$ 's, we limit the variation of other $x_{j}$ 's: treat all $x_{j}$ where $j \neq i$ as constants

{lim}_{Δ x_{i} \to 0} | \frac{f (x_{1}, \dots, x_{i - 1}, x_{i} + Δ x_{i}, x_{i + 1}, \dots) - f (x_{1}, \dots, x_{i - 1}, x_{i}, x_{i + 1}, \dots)}{Δ x_{i}} - α_{i} | = 0 .

Denote $α_{i} = \frac{\partial f}{\partial x_{i}}$ the partial derivatives of function $f$ . Of course, when there is only one variable, partial derivative is the derivative high school textbooks know well.

With this kind of understanding, chain rule is much more intuitive than the otherwise algebraic coincidence.

Product Rule

Now that product rule is no longer a rule, but a natural result of function $f (u, v) = u v$ . Let's assume that $u (t), v (t)$ are functions of $t$ , then by chain rule:

\frac{d f}{d t} = \frac{\partial_{u} f d u + \partial_{v} f d f}{d t},

where $\partial_{u} f$ is an abbreviation of $\frac{\partial f}{\partial u}$ . But obviously, $\partial_{u} f = v$ and $\partial_{v} f = u$ , thus

\frac{d u v}{d t} = u \frac{d v}{d t} + v \frac{d u}{d t} .

An interesting generalisations is

Product Rule with More Multiplicands

Let $f = \prod_{i} x_{i}$ where $x_{i} (t)$ 's are functions of $t$ . By chain rule it is not difficult to prove

\frac{d \prod_{i} x_{i}}{d t} = \prod_{j} x_{j} (\sum_{i} \frac{x_{i}^{'}}{x_{i}}) .

Implicit Differentiation

There is a particular chapter that dedicates to the differentiation of implicit expression. But this becomes unnecessary if we see the expression as multi-variable functions constrained by an equation.

That is,

F (x, y) = 0

Apply our "chain rule" again

\partial_{x} F \times \frac{d x}{d t} + \partial_{y} F \times \frac{d y}{d t} = 0 .

But notice, the terminal variable $t$ is actually $x$ here, therefore

\frac{d y}{d x} = - \frac{\partial_{x} F}{\partial_{y} F} .

This method also applies to the corresponding section in 'Polar Coordinates' where kids are asked to find the the 'gradient' of a curve at some point. Given

F (r, θ) = 0

and $x (r, θ) = r \cos θ, y (r, θ) = r \sin θ$ . By chain rule

d F = \partial_{r} F d r + \partial_{θ} F d θ = \partial_{r} F (\partial_{x} r d x + \partial_{y} r d y) + \partial_{θ} F (\partial_{x} θ d x + \partial_{y} θ d y) = 0 .

We finally obtain

\frac{d y}{d x} = - \frac{\partial_{r} F \partial_{x} r + \partial_{θ} F \partial_{x} θ}{\partial_{r} F \partial_{y} r + \partial_{θ} F \partial_{y} θ} .

Differential Equation

In multi-variable calculus, an important result is Green theorem which stats

\oint_{\partial D} L d x + M d y = \iint_{D} (\partial_{x} M - \partial_{y} L) d x d y .

Therefore, we have such understanding that if we can write a differential equation (with $M, L$ known) in the form

\iint_{D} (\partial_{x} M - \partial_{y} L) d x d y = 0

then the solution would be $\partial_{x} M = \partial_{y} L$ .

I encountered this STEP 2 question the other day:

Solve differential equation

$$ \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{2x-y}{x+4y}. $$

Now this is otherwise a difficult question of high school kids, but if one can see that the D.E. is equivalent to

4 y d y + x d y + y d x - 2 x d x = d (2 y^{2} + x y - x^{2}) = 0 .

It is rather easy to get $2 y^{2} + x y - x^{2} = C$ . Considering the similarity, I believe it is possible to introduce integrating factors in this way.

#"hs-maths"