Implicit differentiation

If a relation can easily be solved as y = f(x), y is an explicit function of x. If not, as in x + sin xy − y = 0, y is given implicitly, yet it still depends on x.

Method: differentiate every term of the relation with respect to x, treating y as a function of x. Each term in y picks up a factor dy/dx by the chain rule; then solve for dy/dx.

Worked example: x − y = π. Differentiating gives 1 − dy/dx = 0, so dy/dx = 1, the same as from the explicit form y = x − π.

Worked example: y + sin y = cos x. Differentiating gives dy/dx + cos y · dy/dx = −sin x, so dy/dx = −sin x/(1 + cos y), wherever y ≠ (2n + 1)π.

Products that contain y need the product rule: d/dx(xy) = y + x dy/dx.

Worked example: x² + xy + y² = 100 gives 2x + y + x y′ + 2y y′ = 0, so y′ = −(2x + y)/(x + 2y).

The answer is allowed to contain both x and y. At a given point on the curve, substitute both coordinates.