Thm 2.3 (Chain rule): Suppose U ⊂ R^{m} is open and f : U → V ⊂ R^{m}, g : V → R^{p}. Let h = g ∘ f. Suppose f is differentiable at a ∈ U and g is differentiable at f(a) ∈ V. Then h is differentiable at a ∈ U and D(h)_{a} = D(G)_{f(a)}D(f)_{a}. Let R_{h}(x, a) = \frac{g(f(x))−g(f(a))−D(g)_{f(a)}D(f)_{a}(x−a)}{||x−a||} Let y = f(x), b = f(a) R_{g}(y, b) = \frac{g(y)−g(b)−D(g)_{b}(y−b)}{||y−b||} where lim_{x→a}R_{g}(y, b) = 0 R_{f}(x, a) = \frac{f(x)−f(a)−D(f)_{a}(x−a)}{||x−a||} where lim_{x→a}R_{f}(x, a) = 0 y − b = f(x) − f(a) = D(f)_{a}(x − a) + ||x − a||R_{f}(x, a) R_{g}(y, b) = \frac{g(f(x))−g(f(a))−D(g)_{b}[D(f)_{a}(x−a)+||x−a||R_{f} (x,a)]}{||y−b||} \frac{||y−b||R_{g}(y,b)}{||x−a||} = \frac{g(f(x))−g(f(a))−D(g)_{b}D(f)_{a}(x−a)−D(g)_{b}||x−a||R_{f} (x,a)}{||x−a||} \frac{||y−b||R_{g}(y,b)+D(g)_{b}||x−a||R_{f} (x,a)}{||x−a||} = \frac{g(f(x))−g(f(a))−D(g)_{b}D(f)_{a}(x−a)}{||x−a||} R_{h}(x, a) = \frac{||y−b||R_{g}(y,b)+D(g)_{b}||x−a||R_{f} (x,a)}{||x−a||} R_{h}(x, a) = \frac{||f(x)−f(a)||R_{g}(y,b)}{||x−a||} + D(g)_{b}R_{f}(x, a) R_{h}(x, a) = \frac{||D(f)_{a}(x−a)+||x−a||R_{f} (x,a)||R_{g}(y,b)}{||x−a||} + D(g)_{b}R_{f}(x, a)

Cor 2.4: If f, g ∈ C^{r} on U, V respectively, then h = g ∘ f ∈ C^{r}. Proof by induction: r = 1: Suppose f, g ∈ C^{1} on U, V respectively. Then \frac{∂f}{∂x_{i}} exists and is continuous on U Then \frac{∂g}{∂x_{i}} exists and is continuous on V. By Thm 1.3, f, g are differentiable on U, V respectively. Hence by Thm 1.1, f is continuous. By Thm 2.3 h = g ∘ f is differentiable. Thus by Thm 1.1, \frac{∂h}{∂x_{i}} exist. By Thm 2.3 it’s Jacobian is D(h)_{x} = D(g)_{f(x)}D(f)_{x}. Since \frac{∂f_{i}}{∂x_{j}} are continuous on U for all i, j, each entry of D(f)_{x} = (\frac{∂f_{i}}{∂x_{j}}) is continuous Since \frac{∂g_{i}}{∂x_{j}} are continuous on V for all i, j and f is continuous, each entry of D(g)_{f(x)} = (\frac{∂g_{i}}{∂x_{j}} |_{f(x)}) is continuous. Since the sums and products of continuous functions are continuous, each entry of D(h)_{a} = D(g)_{f(a)}D(f)_{a} is continuous.