Higher Derivatives of Implicit Functions

The Setup - Slopes We Already Know

Let's start with something familiar. We have a surface defined implicitly - say the sphere $x^2 + y^2 + z^2 = 9$. We can't easily write $z = \ldots$ in a neat formula (well, for a sphere we can, but imagine a more complicated surface where we genuinely can't).

From §2.10, we already know how to find first-order slopes. The key trick: differentiate the whole equation $F(x, y, z(x,y)) = 0$ with respect to $x$, then solve for $z_x$. This gives us:

Try it on the sphere below. Drag the sliders to move along the surface - the orange tangent plane shows the local slope at each point. Notice how the plane tilts more steeply as you approach the equator.

Differentiating Once - Where the Formula Comes From

Before we go to second derivatives, let's make sure we really understand the first. It turns out there's a lovely little mechanism behind that $-F_x/F_z$ formula.

We start from $F(x, y, z(x,y)) = 0$. This is true for all $x$ and $y$ near our point. So we can differentiate both sides with respect to $x$ (holding $y$ constant). The chain rule gives us three contributions - press here and watch them light up:

Two sources of change when we nudge $x$: the direct effect ($F_x$) and the indirect effect through $z$ ($F_z \cdot z_x$). Solve for $z_x$ and you get our formula:

Why the minus sign? Think about it: if increasing $x$ pushes $F$ up ($F_x > 0$), then $z$ has to decrease to keep $F = 0$. The surface compensates.

Differentiating Twice - The Big Formula

Now here's where it gets interesting. We already have the equation $F_x + F_z \cdot z_x = 0$. What if we differentiate this again with respect to $x$?

This is the most genius step: every term in $F_x + F_z \cdot z_x = 0$ is itself a function of $x$ (both directly and through $z$), so we need the chain rule again - and for the $F_z \cdot z_x$ piece, the product rule too. It's a bit of a chain-rule explosion:

I'll be honest - this looks intimidating at first glance. But notice: $z_{xx}$ appears in exactly one place (the last term). Everything else involves $F$-partials and $z_x$, which we already know. So we solve for $z_{xx}$ and substitute $z_x = -F_x/F_z$:

Don't memorize this - understand its three pieces:

• $F_{xx} F_z^2$ - how $F$ curves directly in $x$
• $2F_{xz} F_x F_z$ - the interaction between $x$ and $z$ directions
• $F_{zz} F_x^2$ - how $F$ curves in $z$, weighted by how much $z$ responds to $x$

Mixed Partials - Does the Order Matter?

What if we differentiate $z_x$ with respect to $y$ instead of $x$? We'd get the mixed partial $z_{xy}$. And we could also start from $z_y$ and differentiate with respect to $x$ to get $z_{yx}$.

A natural question: should we expect $z_{xy} = z_{yx}$? Clairaut's theorem says yes - as long as $F$ has continuous second partials, the order doesn't matter. But here's a nice sanity check: let's derive both and watch them match.

The mixed partial comes from differentiating $z_x = -F_x/F_z$ with respect to $y$. Using the quotient rule (remembering that $F_x$ and $F_z$ both depend on $y$ through $z$):

The trap: It's tempting to forget that $F_x$ depends on $y$ through $z$. When you differentiate $F_x$ with respect to $y$, you need $F_{xy} + F_{xz} \cdot z_y$, not just $F_{xy}$. This is the single most common mistake on exams.

Try it on the ellipsoid below. Move the point around - you'll see $z_{xy}$ and $z_{yx}$ always agree, confirming Clairaut's theorem in action.

A Harmonic Function Hiding in Plain Sight

The function $f = \arctan(y/x)$ is the angle in polar coordinates - the $\theta$ you've been drawing since trig class. It has a singularity at the origin (the angle isn't defined there). But away from the origin, is it harmonic?

Predict before computing: the angle function $\theta = \arctan(y/x)$ is smooth and well-behaved everywhere except the origin. Does its Laplacian vanish?

Let's find out. We need $f_{xx} + f_{yy}$ and hope it equals zero.

Step 1: First partials. Using the chain rule with $u = y/x$:

Step 2: Second partials. Let $r^2 = x^2+y^2$. Differentiating $f_x = -y(x^2+y^2)^{-1}$ with respect to $x$:

And differentiating $f_y = x(x^2+y^2)^{-1}$ with respect to $y$:

Step 3: The Laplacian.

The cancellation is exact. And there's a deep reason: $\arctan(y/x)$ is the imaginary part of $\log(x+iy)$, and every analytic function's real and imaginary parts are harmonic. We keep seeing this pattern - the complex-analytic connection is no coincidence.

The Biharmonic Equation - Applying the Laplacian Twice

Every harmonic function solves $\nabla^2 f = 0$. What if we apply the Laplacian twice - requiring $\nabla^2(\nabla^2 f) = 0$? That's the biharmonic equation, and it governs elasticity and plate bending in engineering.

If $f$ is harmonic, then $\nabla^2 f = 0$. So $\nabla^4 f = \nabla^2(0) = 0$. One line. Done. Every harmonic function is automatically biharmonic.

But the converse is false. Here's a clean counterexample.

Take $f = x^2 + y^2$. Its Laplacian:

So $f = x^2+y^2$ is not harmonic. But now apply the Laplacian again:

The Laplacian of a constant is zero. So $x^2+y^2$ is biharmonic but not harmonic.

Predict: the gap between harmonic and biharmonic is exactly the functions whose Laplacian is a nonzero constant. Can you see why? If $\nabla^2 f = c$ for some constant $c$, then $\nabla^4 f = \nabla^2(c) = 0$, but $\nabla^2 f \neq 0$ whenever $c \neq 0$.

The trap: don't confuse "biharmonic" with "twice harmonic." Biharmonic means $\nabla^4 f = 0$, which is weaker than $\nabla^2 f = 0$. Every harmonic function passes the biharmonic test, but many biharmonic functions fail the harmonic test. The hierarchy goes one way only.