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Implicit Differentiation & Related Rates

Some curves, like a circle x^2 + y^2 = 25, can't be neatly solved for y. With [[implicit-differentiation|implicit differentiation]] you can still find their slope — and the same trick, applied to time, becomes [[related-rates|related rates]]: how a sliding ladder, an inflating balloon, or a rising tide change together.

When you can't solve for y

So far, every function has come tidy: y is written explicitly in terms of x, like y = x^2 or y = sin(x), and you differentiate the right side. But plenty of curves refuse that form. The circle x^2 + y^2 = 25 is the classic case: try to solve for y and you get y = +sqrt(25 - x^2) or y = -sqrt(25 - x^2) — two half-circles, not one clean function. Worse curves, like x^3 + y^3 = 6xy, can't be untangled at all. Yet these curves clearly have a tangent line at each point, so there should be a slope. How do we get it without solving for y?

The key idea is a small shift in attitude. Even though we never wrote y as a formula in x, near most points the curve does behave like some function y(x) — pick a point on the upper half of the circle, and locally y really is a function of x. So we treat y as 'y, which secretly depends on x' and differentiate the whole equation. The one rule that makes this work is the chain rule from the previous rung: whenever you differentiate something built out of y, you must multiply by dy/dx, because y itself is changing as x changes.

The implicit recipe, on a circle

Let's find the slope of the circle x^2 + y^2 = 25 at the point (3, 4) — a point that genuinely sits on it, since 9 + 16 = 25. The plan: differentiate both sides with respect to x, remembering that y carries a hidden dy/dx, then solve the resulting equation for dy/dx.

  1. Differentiate every term with respect to x. The x^2 gives 2x; the y^2 gives 2y dy/dx (chain rule); the constant 25 gives 0. Result: 2x + 2y dy/dx = 0.
  2. Solve for dy/dx algebraically: move 2x over and divide by 2y, giving dy/dx = -2x / (2y) = -x / y. Notice the slope depends on both x and y — natural, since each point of the curve has its own tangent.
  3. Plug in the point (3, 4): dy/dx = -3 / 4. The tangent line at (3, 4) slopes gently downhill, which matches the picture of the upper-right of the circle.
x^2 + y^2 = 25

d/dx :  2x + 2y (dy/dx) = 0

        dy/dx = -x / y

at (3, 4):  dy/dx = -3/4
Implicit differentiation: differentiate term by term, attach dy/dx to anything in y, then solve for dy/dx.

Related rates: differentiate with respect to time

Here is where implicit differentiation earns its keep in the real world. Often two or more quantities are tied together by an equation and all of them are changing as time passes. Related rates problems ask: given how fast one quantity changes, how fast does another change? The trick is identical to before — differentiate the relation — except now the hidden variable everything depends on is time t, so every term picks up a d/dt and a chain-rule factor.

Take the favorite example: a 10-foot ladder leans against a wall, and its base is pulled away from the wall at 2 ft/s. How fast is the top sliding down when the base is 6 ft out? The wall, floor, and ladder make a right triangle, so by Pythagoras x^2 + y^2 = 10^2, where x is the base distance and y is the height of the top. Both x and y depend on time. Differentiating with respect to t gives 2x (dx/dt) + 2y (dy/dt) = 0 — the same shape as the circle, but now the dy/dx becomes dy/dt and we also get a dx/dt.

  1. Write the relation and differentiate with respect to t: from x^2 + y^2 = 100 we get 2x (dx/dt) + 2y (dy/dt) = 0, i.e. x (dx/dt) + y (dy/dt) = 0.
  2. Fill in the numbers at the instant asked. When x = 6, the height is y = sqrt(100 - 36) = 8, and we are told dx/dt = +2 (the base moves out). Substitute: 6(2) + 8 (dy/dt) = 0.
  3. Solve for the unknown rate: 8 (dy/dt) = -12, so dy/dt = -1.5 ft/s. The minus sign says the top is sliding down at 1.5 ft/s — slower than the base moves out, just as intuition expects.

A second example, and the shared shape

An inflating spherical balloon shows the same machinery. Its volume is V = (4/3) pi r^3, where both V and r grow over time. Differentiate with respect to t — the chain rule turns r^3 into 3r^2 (dr/dt) — giving dV/dt = 4 pi r^2 (dr/dt). If air is pumped in at dV/dt = 100 cubic cm per second, then when the radius is r = 5 cm we can solve 100 = 4 pi (25)(dr/dt), so dr/dt = 1 / pi cm/s. The balloon's skin pushes outward steadily, but the radius creeps slower and slower as the sphere grows, because the same air now has to coat a larger surface.

Step back and notice that both worlds run the same play. Implicit differentiation differentiates a relation in x and y with respect to x, attaching dy/dx to the y-parts. Related rates differentiates a relation in several quantities with respect to time t, attaching a rate like dx/dt or dr/dt to each changing quantity. In both, the engine is the chain rule honestly applied, and the final move is plain algebra: isolate the derivative you want.