Differentiating composite functions
Theorem 1.0 - Let f(x,y) differentiable in open set A of ℝ2 and
x = x(t), y = y(t), t ∈ [a,b] (1.0)
two differentiable functions in the interval a ≤ t ≤ b with (x(t),y(t)) ∈ A, then ∀t ∈ [a,b] there is a unique value f(x(t),y(t)) = F(t); the function F is a composite function, differentiable in [a,b] such that
F'(t) = fx (x(t),y(t)) ⋅ x'(t) + fy (x(t),y(t)) ⋅ y'(t) (1.1)
this relation is also known as the chain rule for differentiating composite maps.
Proof - The increments of the functions x and y are
h = x(t + dt) − x(t); k = y(t + dt) − y(t)
From the differentiability of f, we can write
F(t+dt) − F(t) = f(x + h, y + k) − f(x,y) = fx(x, y) h + fy(x, y) k + o[√(h2 + k2)]
From the differentiability of x and y, follows
h = x'(t) ⋅ dt + o(dt); k = y'(t) ⋅ dt + o(dt)
Thus
F(t+dt) − F(t) = [fx ⋅ x' + fy ⋅ y'] ⋅ dt + o(dt). □
We can think the functions x(t) and y(t) as components of a regular curve.
Proposition 1.0 - Consider a curve γ: I ⊆ ℝ → ℝ2 of differentiable components (x(t),y(t)), together with a differentiable map f: ℝ2 → ℝ. Let h = f ◦ γ: I → ℝ be the composition
h(t) = f(γ(t)) = f(x(t),y(t))
Then it follows from eq. (1.1) that
h'(t) = ∇f(γ(t)) · γ'(t)
Proposition 1.1 - Let f: ℝ2 → ℝ differentiable , and f(x,y) = c, the equation of one of its level set. Suppose that this level set admits a regular parametric representation by the function γ = γ(t). Then setting
h(t) = f(γ)
It results from the definition, f(t) = c, thus h'(t) = 0. Then it follows from eq. (1.1) that
h'(t) = ∇f(γ(t)) · γ'(t)
This relation tells us that the gradient of f and the vector tangent to the level set γ'(t) are orthogonal along a curve level.
Proposition 1.2 - Let u, v ∈ ℝn be given points and suppose f is differentiable at u + t0 v with t0 ∈ ℝ. Then the map
φ(t) = f(u + tv) (1.4)
is differentiable at t0, and
φ' (t0) = ∇f (u + t0v) · v (1.5)
Proof - (1.5) follows from applying the chain rule (1.1) to (1.4) □.