Differentiation

我的原文地址

Chapter 5: Differentiation In this chapter we shall (except in the final section) confine our attention to real functions defined on intervals or segments. This is not just a matter of convenience, since genuine differences appear when we pass from real functions to vector-valued ones. Differentiation of functions defined on $\mathbb R^k$ will be discussed in Chap. 9.

Contents
Contents
 1.  The derivative of a real function
 2.  Mean value theorems
 3.  The continuity of derivatives
 4.  L'Hospital's rule
 5.  Taylor's theorem
 6.  Differentiation of vector-valued functions

1. The derivative of a real function

The following theorem is known as the ``chain rule" for differentiation. It deals with differentiation of composite functions and is probably the most important theorem about derivatives. We shall meet more general versions of it in Chap. 9.

Our next theorem is the basis of many applications of differentiation.

This theorem is often called a generalized mean value theorem; the following special case is usually referred to as ``the" mean value theorem:

It can be seen that a function/may have a derivative $f^\prime$ which exists at every point, but is discontinuous at some point. However, not every function is a derivative. In particular, derivatives which exist at every point of an interval have one important property in common with functions which are continuous on an interval: Intermediate values are assumed. The precise statement follows.

But $f^\prime$ may very well have discontinuities of the second kind.
4. L'Hospital's rule The following theorem is frequently useful in the evaluation of limits.

5. Taylor's theorem

For $n = 1$, this is just the mean value theorem. In general, the theorem shows that $f$ can be approximated by a polynomial of degree $n-1$, and that it allows us to estimate the error, if we know bounds on $|f^{(n)}(x)|$.
6. Differentiation of vector-valued functions The definition of derivative of real functions applies without any change to complex functions $f$ defined on $[a,b]$, and the first two theorem in this chapter, as well as their proofs, remain valid. If $f_1$ and $f_2$ are the real and imaginary parts of $f$, that is, if $f(t)=f_1(t)+if_2(t)$ for $a\leqslant t\leqslant b$, where $f_1(t)$ and $f_2(t)$ are real, then we clearly have $f^\prime (t)=f^\prime _1(t)+if^\prime _2(t)$; also, $f$ is differentiable at $x$ if and only if both $f_1$ and $f_2$ are differentiable at $x$.

Passing to vector-valued functions in general, i.e., to functions $\mathbf{f}$ which map $[a, b]$ into some $\mathbb R^k$, we may still apply the original definition to define $\mathbf f^\prime (x)$. In other words, $\mathbf f^\prime (x)$ is that point of $\mathbb R^k$ (if there is one) for which
$$\lim_{t\to x}\Big|\frac{\mathbf{f}(t)-\mathbf{f}(x)}{t-x}-\mathbf{f}^\prime (x)\Big|=0,$$
and $\mathbf f^\prime$ is a function with values in $\mathbb R^k$.

If $f_1,\dots,f_k$ are the components of $\mathbf{f}$, then $\mathbf{f}^\prime =(f^\prime _1,\dots,f^\prime _k)$, and $\mathbf f$ is differentiable at a point $x$ if and only if each of the functions $f_1,\dots,f_k$ is differentiable at $x$.

Theorem theorem511 is true in this context as well, and so is Theorem theorem512(1) and (2), if $fg$ is replaced by the inner product $\mathbf{f}\cdot\mathbf{g}$.

When we turn to the mean value theorem, however, and to L^\prime Hospital^\prime s rule, one of its consequences, the situation changes. The next two examples will show that each of these. results fails to be true for complex-valued functions.

Example. Define, for real $x$, $f(x)=e^{ix}=\cos x+i\sin x$. (The last expression may be taken as the definition of the complex exponential $e^{ix}$; see Chap. 8 for a full discussion of these functions.) Then $f(2\pi)-f(0)=0$, but $f^\prime (x)=ie^{ix}$, so that $|f^\prime (x)|=1$ for all real $x$. Thus Theorem theorem521 fails to hold in this case.

Example. On the segment $(0,1)$, define $f(x)=x$, and $g(x)=x+x^2e^{i/x^2}$. We see that $\lim_{x\to0}\frac{f(x)}{g(x)}=1$, while $\lim f^\prime (x)/g^\prime (x)=0$. L^\prime Hospital^\prime s rule fails in this case. Note also that $g^\prime (x)\neq0$ on $(0,1)$.

However, there is a consequence of the mean value theorem which, for purposes of applications, is almost as useful as Theorem theorem523, and which remains true for vector-valued functions: From Theorem theorem523 it follows that
$$|f(b)-f(a)|\leqslant(b-a)\sup_{a<x<b}|f^\prime (x)|.$$

Proof. Put $\mathbf z=\mathbf{f}(b)-\mathbf{f}(a)$, and define $\varphi(t)=\mathbf{z}\cdot\mathbf{f}(t)\, (a\leqslant t\leqslant b)$. Then $\varphi(t)$ is a real-valued continuous function on $[a, b]$ which is differentiable in $(a, b)$. The mean value theorem shows therefore that
$$\varphi(b)-\varphi(a)=(b-a)\varphi^\prime (x)=(b-a)\mathbf z\cdot\mathbf{f}^\prime (x)$$
for some $x\in(a,b)$. On the other hand,
$$\varphi(b)-\varphi(a)=\mathbf z\cdot\mathbf{f}(b)-\mathbf z\cdot\mathbf{f}(a)=|\mathbf{z}|^2.$$

The Schwarz inequality now gives
$$\begin{gather*} |\mathbf{z}|^2=(b-a)|\mathbf z\cdot\mathbf{f}^\prime (x)|\leqslant(b-a)|\mathbf z||\mathbf{f}^\prime (x)|. \ |\mathbf{z}|\leqslant(b-a)|\mathbf{f}^\prime (x)|. \end{gather*}$$
We obtain the desired conclusion.