The Chain Rule, Inverse Functions, Higher Derivatives & The Leibniz Rule

1. Statement of the Chain Rule Theorem

2. The Classic Proof Pitfall & Carathéodory's Resolution

The naive attempt writes $\frac{\Delta y}{\Delta x} = \frac{\Delta y}{\Delta u} \cdot \frac{\Delta u}{\Delta x}$. This fails whenever $\Delta u = g(x_0 + h) - g(x_0) = 0$ for values of $h \ne 0$ (division by zero).

Carathéodory's Theorem: A function $f$ is differentiable at $u_0$ if and only if there exists a function $\Phi(u)$ that is continuous at $u_0$ such that for all $u$:

Rigorous Proof of Chain Rule: Let $u = g(x)$ and $u_0 = g(x_0)$. By Carathéodory's characterization for $f$ at $u_0$:

Dividing by $x - x_0$ for $x \ne x_0$:

As $x \to x_0$: because $g$ is differentiable at $x_0$, it is continuous at $x_0$, so $\lim_{x \to x_0} g(x) = g(x_0) = u_0$. Since $\Phi$ is continuous at $u_0$, $\lim_{x \to x_0} \Phi(g(x)) = \Phi(u_0) = f'(g(x_0))$. Thus:

1. Implicit Functions & The Method of Implicit Differentiation

An equation of the form $F(x, y) = 0$ defines $y$ as an implicit function of $x$. Instead of solving explicitly for $y = f(x)$, we differentiate both sides of $F(x, y) = 0$ with respect to $x$, treating $y$ as an unknown differentiable function of $x$ and applying the Chain Rule $\frac{d}{dx}[g(y)] = g'(y) \frac{dy}{dx}$:

2. Higher-Order Implicit Derivatives

To find the second derivative $\frac{d^2y}{dx^2}$, differentiate the expression for $\frac{dy}{dx}$ with respect to $x$, applying the Quotient Rule, Product Rule, and Chain Rule, and then substitute the expression for $\frac{dy}{dx}$ to express $\frac{d^2y}{dx^2}$ purely in terms of $x$ and $y$.

1. Derivative of the Inverse Function Theorem

2. Derivatives of Inverse Trigonometric Functions

Let $y = \arcsin x$ for $x \in (-1, 1)$. Then $x = \sin y$ with $y \in (-\pi/2, \pi/2)$:

Similarly:

3. Derivatives of Inverse Hyperbolic Functions

1. Logarithmic Differentiation

For functions involving complicated products, quotients, or variable powers $y = [u(x)]^{v(x)}$:

1. Take the natural logarithm of both sides: $\ln y = v(x) \ln u(x)$.
2. Differentiate implicitly with respect to $x$: $\frac{1}{y} \frac{dy}{dx} = v'(x) \ln u(x) + v(x) \frac{u'(x)}{u(x)}$.
3. Multiply through by $y$: $\frac{dy}{dx} = [u(x)]^{v(x)} \left[ v'(x) \ln u(x) + \frac{v(x) u'(x)}{u(x)} \right]$.

2. Successive Derivatives & Notations

The $n$-th derivative $f^{(n)}(x) = \frac{d^n y}{dx^n}$ is obtained by iteratively differentiating $n$ times.

3. The General Leibniz Rule for the $n$-th Derivative of a Product

where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ are binomial coefficients. This theorem is proven rigorously by mathematical induction using Pascal's identity $\binom{n}{k-1} + \binom{n}{k} = \binom{n+1}{k}$.