The Timpson Lyceum · Real Analysis (2020)

Lecture 16 (Real Analysis 2020)

Overview

This lecture covers three related ideas. First is the Cauchy Mean Value Theorem, a generalization of the ordinary Mean Value Theorem to a pair of functions, proved using the same Rolle's-theorem trick. Second is a Cauchy-type remainder theorem (Theorem 4.24): by applying the Cauchy MVT repeatedly to a function whose first several derivatives vanish at a point, we extract a single point $c$ at which an $n$th derivative captures the function's value — the machinery behind Taylor-series approximation. Finally, the lecture introduces Leibniz notation $\frac{dy}{dx}$, reinterpreting the derivative as a limit of $\frac{\Delta y}{\Delta x}$ and showing why treating $dy$ and $dx$ like a ratio makes inverse-function derivatives and implicit differentiation easy.

Key Concepts

Cauchy Mean Value Theorem (Thm 4.23). Let $f, g : [a,b] \to \mathbb{R}$ be continuous on $[a,b]$ and differentiable on $(a,b)$, with $g'(x) \neq 0$ for all $x \in (a,b)$. Then there exists $x_0 \in (a,b)$ such that $\frac{f'(x_0)}{g'(x_0)} = \frac{f(b) - f(a)}{g(b) - g(a)}.$

• Interpretation. The left side is the ratio of the two slopes at a single point; the right side is the ratio of the two functions' average slopes over $[a,b]$ (the common $\frac{1}{b-a}$ cancels).
• The ordinary MVT is a special case. Taking $g(x) = x$ gives $g'(x) = 1$, $g(b) - g(a) = b - a$, which collapses the statement to $f'(x_0) = \frac{f(b)-f(a)}{b-a}$.

Rolle's Theorem (the engine of the proof). If $h$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $h(a) = h(b)$, then $h'(x_0) = 0$ for some $x_0 \in (a,b)$.

Auxiliary function technique. To prove the Cauchy MVT, define $h(x) = f(x) - m\,g(x), \qquad m = \frac{f(b)-f(a)}{g(b)-g(a)}.$ One checks $h(a) = h(b)$, applies Rolle's theorem to get $h'(x_0) = 0$, and since $h'(x) = f'(x) - m\,g'(x)$, solving gives $\frac{f'(x_0)}{g'(x_0)} = m$ (legal because $g'(x_0) \neq 0$).

A function that "kills" its low-order derivatives. For $g(x) = (x - x_0)^n$, repeated differentiation gives $g^{(k)}(x_0) = 0$ for $0 \le k < n$, while $g^{(n)}(x_0) = n!$. (Example: $g(x) = (x-a)^4 \Rightarrow g'= 4(x-a)^3,\ g''=12(x-a)^2,\ g'''=24(x-a),\ g^{(4)}=24=4!$.) This is the model behavior assumed in the next theorem.

Cauchy remainder theorem (Thm 4.24). Let $I$ be an open interval, $n \in \mathbb{N}$, and $f : I \to \mathbb{R}$ have $n$ derivatives, with $f^{(k)}(x_0) = 0$ for all $0 \le k < n$ at some $x_0 \in I$. Then for each $x \neq x_0$ in $I$ there is a point $c$ between $x_0$ and $x$ such that $f(x) = \frac{f^{(n)}(c)}{n!}\,(x - x_0)^n.$

• Why it matters. This is the seed of the Taylor series: it lets messy functions (e.g. $\sin x$) be approximated by polynomials, which is how such functions are evaluated in practice.
• Proof idea. Set $g(x) = (x - x_0)^n$ and apply the Cauchy MVT repeatedly. Each application bumps the order of differentiation by one while the quotient stays equal to $\frac{f(x)}{g(x)}$ (because both functions and their low-order derivatives vanish at $x_0$). After $n$ steps you reach $\frac{f^{(n)}(c)}{g^{(n)}(c)} = \frac{f(x)}{g(x)}$, and $g^{(n)}(c) = n!$.

Leibniz notation. Starting from the difference quotient with $\Delta x = x - x_0$ and $\Delta y = f(x) - f(x_0)$, $\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}.$ This is the same derivative as $f'$, with $x$ now playing the role the base point $x_0$ played before. The symbol $\frac{dy}{dx}$ ("derivative of $y$ with respect to $x$") is not literally a fraction, but in many situations it can be manipulated as if $dy$ and $dx$ were separate quantities.

Inverse-function derivative. Since $f^{-1}$ undoes $f$, $\big(f^{-1}\big)'(y_0) = \frac{1}{f'(x_0)}, \qquad \text{i.e.} \qquad \frac{dx}{dy} = \frac{1}{\,dy/dx\,}.$ Reading $\frac{dx}{dy}$ as "$x$ as a function of $y$" makes the reciprocal relationship transparent.

Implicit differentiation. When $y$ is defined implicitly by an equation in $x$ and $y$, differentiate both sides with respect to $x$, treating $y$ as a function of $x$ (so every $y$-term carries a chain-rule factor $\frac{dy}{dx}$, while $\frac{dx}{dx} = 1$), then solve algebraically for $\frac{dy}{dx}$.

Worked Example

Implicit differentiation of a circle. Differentiate $x^2 + y^2 = 1$ with respect to $x$ and solve for the slope $\frac{dy}{dx}$.

Check. At the top of the circle $(x, y) = (0, 1)$ the formula gives $\frac{dy}{dx} = -\frac{0}{1} = 0$, matching the horizontal tangent there. The expression is undefined where $y = 0$ (the points $(\pm 1, 0)$), exactly where the circle has a vertical tangent — consistent with the curve splitting into the two branches $y = \pm\sqrt{1 - x^2}$ on $[-1, 1]$.

Summary

• The Cauchy Mean Value Theorem equates the ratio of two functions' slopes at one interior point to the ratio of their average slopes; the ordinary MVT is the case $g(x) = x$.
• It is proved exactly like the MVT: build $h(x) = f(x) - m\,g(x)$, show $h(a) = h(b)$, and apply Rolle's theorem.
• A function of the form $(x - x_0)^n$ has all derivatives up to order $n-1$ vanishing at $x_0$, with the $n$th derivative equal to $n!$.
• Theorem 4.24 applies the Cauchy MVT repeatedly to such functions, yielding $f(x) = \frac{f^{(n)}(c)}{n!}(x - x_0)^n$ — the foundation of Taylor-polynomial approximation.
• Leibniz notation $\frac{dy}{dx}$ rewrites the derivative as $\lim_{\Delta x \to 0}\frac{\Delta y}{\Delta x}$; though not a true fraction, it can often be manipulated like one.
• This ratio view makes the inverse-function rule $\frac{dx}{dy} = 1\big/\frac{dy}{dx}$ and implicit differentiation (differentiate, attach $\frac{dy}{dx}$ to each $y$-term, solve) immediate.