Higher-Order Derivatives

Definition

If a function $f$ is differentiable everywhere in an interval $I$, then the function

$$\begin{equation*} f^{(1)} := f' : I \to \mathbb{R} \end{equation*}$$

is called the first derivative of $f$. We denote $f^{(0)} := f$.

By induction, if the $n$-th derivative $f^{(n)}$ of $f$ exists, and $f^{(n)}$ is differentiable at $x_0$, then we define

$$\begin{equation*} f^{(n+1)}(x_0) = \big(f^{(n)}\big)'(x_0). \end{equation*}$$

We call $f^{(n+1)}(x_0)$ the $(n+1)$-th derivative of $f$ at $x_0$. The second and third derivatives are denoted by $f''$ and $f'''$, respectively.

We also use the following notations:

$$\begin{equation*} f \in \mathcal{C}^n(I) \quad \text{means } f^{(n)} \text{ is continuous on } I; \end{equation*}$$ $$\begin{equation*} f \in \mathcal{C}^\infty(I) \quad \text{means } f \text{ has derivatives of all orders on } I. \end{equation*}$$

Four Basic Operations

Let $f, g$ have $n$-th derivatives at $x_0$. Then $\lambda f + \mu g$ and $fg$ also have $n$-th derivatives at $x_0$, and we have:

$$\begin{equation*} (\lambda f + \mu g)^{(n)}(x_0) = \lambda f^{(n)}(x_0) + \mu g^{(n)}(x_0) \end{equation*}$$ $$\begin{equation*} (fg)^{(n)}(x_0) = \sum_{k=0}^{n} C_n^k\, f^{(k)}(x_0)\, g^{(n-k)}(x_0) \end{equation*}$$

If $f$ is $n$-times differentiable at $x_0$, and $g$ is $n$-times differentiable at $y_0 = f(x_0)$, then the composite function $g \circ f$ is $n$-times differentiable at $x_0$.

Let $f, g$ be $n$-times differentiable at $x_0$, and suppose $g(x_0) \neq 0$. Then $\dfrac{f}{g}$ is $n$-times differentiable at $x_0$.

The functions $\arctan x$, $\arcsin x$, and $\arccos x$ are all $\mathcal{C}^\infty$ functions. Their derivatives are given by:

$$\begin{equation*} (\arctan x)' = \frac{1}{1 + x^2}, \quad (\arcsin x)' = \frac{1}{\sqrt{1 - x^2}}, \quad (\arccos x)' = \frac{-1}{\sqrt{1 - x^2}}. \end{equation*}$$

Suppose $f$ is $n$-times differentiable on an interval $I$, and $f'(x) \ne 0$. (From continuity, $f$ has a continuous inverse function on $I$.) Then $f^{-1}$ is $n$-times differentiable on the interval $f(I)$.

$$\begin{equation*} \text{If } f \in C^n(I) \text{ and } f'(x) \ne 0, \text{ then } f^{-1} \in C^n(f(I)). \end{equation*}$$

Corollay: All elementary functions are $C^{\infty}$ functions.

Curvature Center, Radius, and Curvature of a Plane Curve

Solution. Let the curve be parameterized as $x = x(t), \, y = y(t)$.

The tangent line of the curve is

$$\begin{equation*} Y - y(t) = \frac{y'(t)}{x'(t)} \,[X - x(t)] \end{equation*}$$

The normal line is

$$\begin{equation*} y'(t) \,[Y - y(t)] = -x'(t) \,[X - x(t)] \end{equation*}$$

At a nearby point, the normal line is

$$\begin{equation*} y'(t+h) \,[Y - y(t+h)] = -x'(t+h) \,[X - x(t+h)] \end{equation*}$$

Expanding by Taylor’s theorem:

$$\begin{equation*} [y'(t) + h y''(t) + o(h)] \,[Y - y(t) - y'(t)h + o(h)] = - [x'(t) + x''(t)h + o(h)] \,[X - x(t) - x'(t)h - o(h)] \end{equation*}$$ $$\begin{equation*} \Rightarrow \; y'(t) \,[Y - y(t)] + x'(t) \,[X - x(t)] + h \,[\,y''(t)(Y - y(t)) + x''(t)(X - x(t)) - (y'(t))^2 - (x'(t))^2\,] + o(h) = 0 \end{equation*}$$

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When $h \to 0$, the intersection of the two normals satisfies:

$$\begin{equation*} \begin{cases} y'(t) \,[Y - y(t)] + x'(t) \,[X - x(t)] = 0, \\ y''(t) \,[Y - y(t)] + x''(t) \,[X - x(t)] = (x'(t))^2 + (y'(t))^2. \end{cases} \end{equation*}$$

Solving for the center of curvature $(X, Y)$:

$$\begin{equation*} \begin{cases} X = x(t) - \dfrac{y'(t) \,\big[(x'(t))^2 + (y'(t))^2\big]} {\begin{vmatrix} x'(t) & x''(t) \\[4pt] y'(t) & y''(t) \end{vmatrix}}, \\[16pt] Y = y(t) + \dfrac{x'(t) \,\big[(x'(t))^2 + (y'(t))^2\big]} {\begin{vmatrix} x'(t) & x''(t) \\[4pt] y'(t) & y''(t) \end{vmatrix}}. \end{cases} \end{equation*}$$

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The radius of curvature is the distance from $(X, Y)$ to $(x(t), y(t))$:

$$\begin{equation*} R = \sqrt{[X - x(t)]^2 + [Y - y(t)]^2} = \frac{\big(\sqrt{(x'(t))^2 + (y'(t))^2}\big)^3} {\Big| \det \begin{pmatrix} x'(t) & x''(t) \\ y'(t) & y''(t) \end{pmatrix} \Big|} \end{equation*}$$

The curvature is the reciprocal of the radius of curvature:

$$\begin{equation*} \kappa = \frac{\Big| \det \begin{pmatrix} x'(t) & x''(t) \\ y'(t) & y''(t) \end{pmatrix} \Big|} {\big[(x'(t))^2 + (y'(t))^2\big]^{3/2}} \end{equation*}$$