🌀Implicit Differentiation

Everything you have differentiated so far arrived already solved for y: y = x² - 3x, y = sin(x), y = e^x. Archmage Newt calls that an explicit equation, because it says outright what y is.

The Spire is built on curves that will not say. A circle is the plainest example:

x² + y² = 25

There is no way to write that as a single y = something. Solve it and the circle splits in two:

y = sqrt(25 - x²) (the top half) y = -sqrt(25 - x²) (the bottom half)

Two functions, two formulas, two derivatives, and a fight with the chain rule and a square root in each of them. An equation like this, one that ties x and y together without solving for either, is called an implicit relation.

You do not have to solve it. Here is the idea the whole craft rests on: wherever you are standing on the curve, y is some function of x, even though nobody has written down which one. Near the point (3, 4) the circle really is the graph of a function; near (3, -4) it is the graph of a different one. Either way there is a y that depends on x, so there is a dy/dx, and the chain rule can reach it.

So differentiate both sides of the equation exactly as it stands, treat every y as "some function of x", and then solve the result for dy/dx. That is implicit differentiation, start to finish.

Differentiating x-terms is nothing new: d/dx of x² is 2x. The whole difference is what happens to a y.

Because y is a function of x, differentiating y² with respect to x is a chain rule: the outside is (something)^2, the inside is y, and the derivative of the inside is dy/dx. Writing y' as a short name for dy/dx:

d/dx (y²) = 2y y' d/dx (y³) = 3y² y' d/dx (sin y) = cos(y) y' d/dx (e^y) = e^y y' d/dx (y) = y'

The pattern never changes. Differentiate as usual, then multiply by y'. A term made only of x picks up no y'; a term with a y in it always does.

A term with both letters needs the product rule as well, and this is where most of the marks are lost:

d/dx (xy) = (1)(y) + (x)(y') = y + x y' d/dx (x² y) = 2xy + x² y' d/dx (x² y³) = 2x y³ + 3x² y² y'

Read the middle one slowly. Half of it, 2xy, came from differentiating the x² and leaving y alone. The other half, x² y', came from differentiating the y and leaving x² alone. Both halves are always there.

The method, in four moves.

• ◆Differentiate every term on both sides with respect to x, attaching y' to each term that came from a y.
• ◆Move every term containing y' to one side and everything else to the other.
• ◆Factor y' out of that side.
• ◆Divide, and you have dy/dx.

For the circle:

x² + y² = 25 2x + 2y y' = 0 2y y' = -2x y' = -x/y

Notice the shape of the answer. dy/dx is usually an expression in x and y, not in x alone — and that is not a failure to simplify. The circle has two points above every x-value, one with a positive slope and one with a negative one, and the y in the formula is exactly what tells them apart.

Once you have dy/dx, a point does the rest of the work. Substitute both coordinates.

On x² + y² = 25 the slope formula is y' = -x/y, so:

at (3, 4): slope = -3/4 at (3, -4): slope = 3/4 at (-4, 3): slope = 4/3

Same x, opposite y, opposite slope. The formula needed the y after all.

Check that the point is on the curve before you use it. Substituting (3, 4) into x² + y² gives 9 + 16 = 25, so the point is genuinely on the circle. A point that is not on the curve has no tangent there, and the arithmetic will still produce a number, which is what makes the slip hard to see.

The tangent line. A line needs a slope and a point, and you now have both. Use the point-slope form and tidy:

y = m(x - x1) + y1

At (3, 4) on the circle, m = -3/4:

y = -3/4 (x - 3) + 4 y = -3x/4 + 9/4 + 4 y = -3x/4 + 25/4

The normal line is the line through the same point at right angles to the curve — straight through the centre, for a circle. Perpendicular slopes multiply to -1, so the normal slope is the negative reciprocal of the tangent slope. At (3, 4) the tangent slope is -3/4, so the normal slope is 4/3, and the normal line is

y = 4/3 (x - 3) + 4 = 4x/3

which passes through the origin, exactly as a circle's normal must.

Write dy/dx as a fraction and the two special directions are easy to spot.

• ◆The tangent is horizontal where dy/dx = 0, which happens where the numerator is zero (and the denominator is not).
• ◆The tangent is vertical where dy/dx has no value, which happens where the denominator is zero (and the numerator is not).

On x² + y² = 25, dy/dx = -x/y. The numerator is zero when x = 0, and the curve then gives y² = 25, so y = 5 or y = -5: the tangent is horizontal at (0, 5) and (0, -5), the top and the bottom of the circle. The denominator is zero when y = 0, giving x = 5 or x = -5: the tangent is vertical at (5, 0) and (-5, 0), the two sides. Four points in all, which is what a circle should have.

The condition is only half the job. Always substitute the condition back into the original equation to find the other coordinate — the answer is a point, not a value of x.

On a curve with a cross term the condition is more interesting. For x² + xy + y² = 12,

2x + y + x y' + 2y y' = 0 y' = -(2x + y)/(x + 2y)

so the tangent is horizontal where 2x + y = 0, that is y = -2x. Substituting into the curve: x² - 2x² + 4x² = 3x² = 12, so x = 2 or x = -2, and the points are (2, -4) and (-2, 4).

The second derivative. Differentiate dy/dx again, still treating y as a function of x, and then substitute the dy/dx you already know. For the circle:

y' = -x/y y'' = -(y - x y')/y² (quotient rule) y'' = -(y + x²/y)/y² (using y' = -x/y) y'' = -(x² + y²)/y³ (multiplying top and bottom by y)

Every second derivative from an implicit relation is found this way: differentiate the first derivative, then replace the y' that appears with the expression you already have. On a circle of radius 5 the answer is also -25/y³, since x² + y² = 25 there — but the form in x and y is the one that is true whichever circle you are on, so that is the one to write down.