Implicit Differentiation, What's Going On Here?

6Implicit Differentiation, What's Going On Here?

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This is Chapter 6 of the Essence of Calculus series. Up to now every derivative had a clean setup: y = f(x), one input, one output, take the slope. This chapter breaks that comfort. We look at curves where y is tangled up with x, and ask how the slope can still exist.

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This series follows 3Blue1Brown's "Essence of Calculus". Watch Chapter 6 here: Implicit differentiation, what's going on here?

The circle that breaks the rules

Take the circle of radius 5, centered at the origin:

x² + y² = 5²

Pick a point on it, say (3, 4). The circle clearly has a tangent line there, so it has a slope. But try to write y as a function of x and you hit a wall. Solving gives y = ±√(25 - x²). That ± is the problem: above the x-axis is one function, below it is another, and at the far left and right edges the tangent is vertical, so the slope is not even a number.

        y
        ^
      • | •
   •    |    •   ← tangent at (3,4)
  •     |     ⟍
 •------+------•----> x
  •     |     •
   •    |    •
      • | •

So dy/dx for the whole circle at once does not fit the "one input, one output" mold. We need a way to find the slope without ever solving for y.

Definition

A way to find a slope when `y` is tangled into the equation instead of written as a clean function of `x`. You differentiate both sides of the equation with respect to `x`, treating `y` as something that secretly depends on `x`, then solve for `dy/dx`.

The trick, step by step

Here is the move that looks like magic. Take the derivative of both sides of x² + y² = 5² with respect to x. The key: treat y as if it were a function of x, even though we never wrote it that way.

d/dx ( x² + y² ) = d/dx ( 25 )

d/dx (x²) = 2x. Easy, that is just the power rule.
d/dx (y²) = 2y · dy/dx. This is the chain rule. y² is an outer function squaring an inner thing y, and y itself changes as x changes, so we tack on dy/dx.
d/dx (25) = 0. A constant never changes.

Put it together:

2x + 2y · dy/dx = 0

Now solve for the slope:

dy/dx = -x / y

Plug in (3, 4): the slope is -3/4. Plug in a point on the right edge like (5, 0) and you get division by zero, which is exactly the vertical tangent we expected. The formula even tells you where the trick honestly breaks.

Why the dy/dx appears out of nowhere

The dy/dx is not a rabbit pulled from a hat. It is the chain rule being honest. The instant you write d/dx (y²), you are differentiating a function of x (because y rides along on x), so the chain rule demands the extra factor dy/dx. Forgetting it is the single most common mistake here.

The deeper picture: keep the output fixed

The step-by-step recipe works, but it can feel like symbol pushing. There is a cleaner way to see it.

Think of the expression x² + y² as a machine that takes a point (x, y) and spits out a value. The circle is the set of points where that machine outputs exactly 25. Staying on the curve means the output must never change. It is locked at 25.

Now nudge the point by a tiny step (dx, dy) along the curve. How much does the value x² + y² change? Each variable contributes its own little nudge:

nudge from x : 2x · dx
nudge from y : 2y · dy
total nudge  : 2x · dx + 2y · dy

To stay on the curve, the total nudge has to be zero, because the output is not allowed to drift off 25:

2x · dx + 2y · dy = 0

Divide through by dx and you are right back at 2x + 2y · dy/dx = 0. Same answer, but now you can see why. The two terms are competing nudges that must cancel.

This "sum the nudges and force them to cancel" view is the whole idea. It does not care whether you can solve for y. It just asks: which directions of travel leave the value unchanged?

Related rates: the same trick over time

The exact same machinery powers related rates problems, where two quantities are linked by an equation and both change over time.

Classic setup: a 5 meter ladder leans against a wall. Its base slides away from the wall, its top slides down. The base distance x and the height y always satisfy:

x² + y² = 5²

Same equation as the circle. But now x and y both depend on time t, so differentiate both sides with respect to t instead of x:

2x · dx/dt + 2y · dy/dt = 0

If the base slides out at dx/dt = 1 m/s, this lets you solve for how fast the top falls, dy/dt, at any moment. The nudges-must-cancel logic is identical. We only swapped the variable we differentiate against.

The mental model to keep

An equation is a constraint: it pins some expression to a constant value. Differentiating both sides means asking how each variable's tiny nudge contributes to changing that value, then forcing the nudges to cancel so you never leave the curve. Solve for the rate you want.

Why this matters for AI

Where this shows up in machine learning

The mindset here, differentiate both sides, treat every variable as changing, and sum the nudges, is exactly how multivariable gradients work. A gradient is just the full list of "how much does each variable nudge the output". When you train a model, backpropagation is this same nudge accounting run across millions of weights. And the "keep the output fixed while moving along a constraint" picture is precisely the idea behind constrained optimization and Lagrange multipliers: optimize a quantity while staying on a surface defined by an equation. You just met the seed of all of it.

Quick gotchas

The chain rule factor is not optional. d/dx (y²) is 2y · dy/dx, never just 2y. The dy/dx is the entire reason implicit differentiation works. Drop it and every answer is wrong.

The answer can contain both x and y. dy/dx = -x/y depends on where you are on the curve. That is expected. A circle has a different slope at every point, so the slope formula must reference the point.

Vertical tangents show up as division by zero. When y = 0 the formula blows up. That is not a bug, it is the math telling you the tangent is vertical and the slope genuinely does not exist as a number there.

Pick the right variable to differentiate against. For a slope dy/dx, differentiate with respect to x. For a related rates problem where things change over time, differentiate with respect to t. Same technique, different driving variable.

What you walked away with

A curve like x² + y² = 25 has a slope everywhere, even though y is not a clean function of x.
Implicit differentiation: differentiate both sides with respect to x, treat y as depending on x (chain rule gives the dy/dx), then solve.
The deeper view: an equation locks an expression to a constant, so a tiny step along the curve must leave the value unchanged. The nudges 2x·dx + 2y·dy cancel to zero.
Related rates is the identical trick with time as the driving variable.
This nudge-accounting is the on-ramp to multivariable gradients and constrained optimization in machine learning.

Next up, Chapter 7: we confront the limit head on. What does it really mean for a step to get "infinitely small", and how does the formal (epsilon, delta) definition pin that idea down? See you there.