Parametric differentiation

Sometimes x and y are each given in terms of a third variable, a parameter t: x = f(t), y = g(t).

Then dy/dx = (dy/dt)/(dx/dt), provided dx/dt ≠ 0. The answer is in terms of the parameter only.

Worked example: the circle x = a cos θ, y = a sin θ. dx/dθ = −a sin θ and dy/dθ = a cos θ, so dy/dx = −cot θ.

Worked example: the parabola x = at², y = 2at. dx/dt = 2at and dy/dt = 2a, so dy/dx = 1/t.

Worked example: the curve x = a(θ + sin θ), y = a(1 − cos θ) has dy/dx = a sin θ / a(1 + cos θ), which simplifies to tan(θ/2).

A relation can be turned into a parametric pair. For x^(2/3) + y^(2/3) = a^(2/3), put x = a cos³θ, y = a sin³θ; then dy/dx = −tan θ, which equals −(y/x)^(1/3).

One function against another: to differentiate u(x) with respect to v(x), use du/dv = (du/dx)/(dv/dx). For sin²x against e^(cos x) this is 2 sin x cos x divided by −sin x e^(cos x), giving −2 cos x/e^(cos x).