Implicit Differentiation CalcLens

Type an equation in x and y and get dy/dx, one move at a time. The whole difficulty is a single step: y is not a free letter. It is whatever function of x the equation forces, so every y you differentiate picks up a y′. Say what the next line should be before you press the button — the forgotten y′ is the whole exam question. An ordinary y = … works here too, and stops as soon as it is solved.

Find d d

Both sides, with an = between them. Write it as you would say it: x^2 + y^2 = 25, x^3 + y^3 = 6xy, sin(y) + x^2 = y, xy = 1.

The working

Implicit Differentiation

Some curves are not the graph of a function. The circle fails the vertical line test, so you cannot write y = something and differentiate it. But the equation still forces a relationship, and near any point on it y is a function of x — just one you never write down.

The one step that matters

Because y is a function of x, the chain rule applies to it:

Writing instead is the single most common error on this topic, and it is the only thing that makes implicit differentiation different from the differentiation you already know. Everything after it is algebra.

Any two letters

The box says which derivative to find, and it is filled in from the equation: V = (4/3)πr³ opens as dV/dr, s² + t² = 9 as ds/dt. Change either letter to ask a different question of the same equation.

What this cannot do is related rates, where several letters all depend on t at once — dV/dt = 4πr²·dr/dt has two dependent quantities, and this page handles one.

It handles an ordinary function too

Put in y = 3x³ + sin x and the working stops the moment y′ stands alone — there is nothing to gather, nothing to factor and nothing to divide by. That is not a special case bolted on; it is the point. Implicit differentiation contains the ordinary kind. It is one procedure, and the extra moves only appear when the equation actually needs them.

Reading the marks

A rule is written above the operator it put there, the way it goes on a board: PR over the + a product rule introduces, QR over the − a quotient rule introduces, CR over the · a chain rule introduces. The operator is the rule's signature, so the mark lands exactly where the rule acted — which matters most on a line where two different rules fire on the two sides at once.

The four moves

  1. Differentiate both sides with respect to x.
  2. Apply the usual rules — and put a y′ on every y term.
  3. Gather the y′ terms on one side, everything else on the other.
  4. Divide by whatever multiplies y′.

Step 3 works because y′ only ever appears to the first power — the chain rule can multiply by it once and no more — so the equation is always linear in y′ and always solvable.

Why the answer has a y in it

That is not a mistake and it does not need "fixing". It is the honest answer: on the circle, the slope at x = 3 is −3/4 on the top half and +3/4 on the bottom. Without saying which y you mean, there is no one slope. Use the box underneath to put in a point — it will tell you if the point is not on the curve, because a slope at a point that is not there means nothing.

Where else to look

For the ordinary rules applied one at a time, use Worked Derivative.

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