4.2 The derivative of a function  (Page 2/4)

Set

Then clearly

which is [link] . Also

which tends to 0 as $h$ approaches 0 because $f$ is differentiable at $c.$ Hence, we have established [link] and [link] , showing that (1) implies (3).

Finally, suppose there is a number $L$ and a function $\theta$ satisfying [link] and [link] . Then

which converges to $L$ as $h$ approaches 0 by [link] and part (2) of [link] . Hence, $L=f^{\text{'}}\left(c\right),$ and so $\theta \left(h\right)=f(c+h)-f\left(c\right)-f^{\text{'}}\left(c\right)h.$ Therefore, (3) implies (1), and the theorem is proved.

REMARK Though it seems artificial and awkward, Condition (3) of this theorem is very convenient for many proofs.One should remember it.

If $f:S\to C$ is a function, either of a real variable or a complex variable, and if $f$ is differentiable at a point $c$ of $S,$ then $f$ is continuous at $c.$ That is, differentiability implies continuity.

We are assuming that ${lim}_{h\to 0}(f(c+h)-f\left(c\right))/h=L.$ Hence, there exists a positive number ${\delta }_0$ such that $|\frac{f(c+h)-f\left(c\right)}{h}-L|<1$ if $\left|h\right|<{\delta }_0,$ implying that $\left|f\right(c+h)-f(c\left)\right|<\left|h\right|\left(\right|L|+1)$ whenever $\left|h\right|<{\delta }_0.$ So, if $ϵ>0$ is given, let $\delta$ be the minimum of ${\delta }_0$ and $ϵ/\left(\right|L|+1).$ If $y\in S$ and $|y-c|<\delta ,$ then, thinking of $y$ as being $c+h,$

(Every $y$ can be written as $c+h$ for some $h,$ and $|y-c|=\left|h\right|.$ )

The following theorem generalizes the preceding exercise.

Suppose $f:S\to R$ is a real-valued function of a complex variable,and assume that $f$ is differentiable at a point $c\in S.$ Then $f^{\text{'}}\left(c\right)=0.$ That is, every real-valued, differentiable function $f$ of a complex variable satisfies $f^{\text{'}}\left(c\right)=0$ for all $c$ in the domain of $f^{\text{'}}.$

We compute $f^{\text{'}}\left(c\right)$ in two ways.

Hence, $f^{\text{'}}\left(c\right)$ must be 0, as claimed.

REMARK This theorem may come as a surprise, for it shows that there are very few real-valued differentiable functions of a complex variable. For this reason, whenever $f:S\to R$ is a real-valued, differentiable function, we will presume that $f$ is a function of a real variable; i.e., that the domain $S\subseteq R.$

Evaluating ${lim}_{h\to 0}q\left(h\right)$ in the two different ways, $h$ real, and $h$ pure imaginary, led to the proof of the last theorem. It also leads us to make definitions of what are called “partial derivatives”of real-valued functions whose domains are subsets of $C\equiv R^2.$ As the next exercise will show, the theory of partial derivatives of real-valued functions is a much richer theory than that of standard derivatives ofreal-valued functions of a single complex variable.