Differential Calculus

Theorem (Chain Rule): Let $F: \mathbf{R} \rightarrow \mathbf{R}$ be differentiable at $u(a)$ and $v: \mathbf{R} \rightarrow \mathbf{R}$ be differentiable at $a$. Then $(F \circ u)'(a) = F'(u(a))\,u'(a)$.

Proof: Since $u$ is differentiable at $a$, we know there must exist some function $\chi$ continuous at $0$ such that $u(x + h) = u(a) + \chi(h)h$. Let $k_h = \chi(h)h$. Then, since $F$ is differentiable at $u(a)$, there must exist some function $\psi$ continuous at $0$ such that $\psi(0) = F'(u(a))$, and: $$\begin{align*} F(u(x + h)) - F(u(a)) &= F(u(a) + \chi(h)h) - F(u(a))\\ &= F(u(a) + k_h) - F(u(a))\\ &= \psi(k_h)k_h = \psi(\chi(h)h)\chi(h)h\\ \end{align*}$$

Since continuous functions are closed under composition and multiplication, $F \circ u$ is differentiable, with $\phi(h) = \psi(\chi(h)h)\chi(h)$ defining the necessary continuous function $\phi$. Thus, $(F \circ u)'(x) = \psi(0)\chi(0) = F'(u(a))u'(a)$. $\square$

Theorem (Product Rule): Let $u$ and $v$ be functions from $\mathbf{R}$ to $\mathbf{R}$ which are differentiable at $a \in \mathbf{R}$. Then $(uv)'(a) = u(a)v'(a) + v(a)u'(a)$.

Proof: Since $u$ and $v$ are differentiable, we know there must exist functions $\chi$ and $\psi$ continuous at $0$ such that $\chi(0) = u'(a)$, $\psi(0) = v'(a)$, and:

$$\begin{align*} (uv)(a+h) - (uv)(a) &= u(a+h)v(a+h) - u(a)v(a)\\ &= (u(a) + \chi(h)h)(v(a) + \psi(h)h) - u(a)v(a)\\ &= u(a)\psi(h)h + v(a)\chi(h)h + \psi(h)\chi(h)h^2\\ &= (u(a)\psi(h) + v(a)\chi(h) + \psi(h)\chi(h)h)h\\ \end{align*}$$

Since continuous functions are closed under scalar multiplication, addition, and multiplication, we see that $uv$ is differentiable, with $\phi(h) = u(a)\psi(h) + v(a)\chi(h) + \psi(h)\chi(h)h$ providing the definition of the necessary continuous function $\phi$. Therefore $(uv)'(x) = \phi(0) = u(a)v'(a) + v(a)u'(a)$. $\square$

Exercises