ANA1_L23

Derivatives of real functions

Rudin, 5.1

Definition 1(Definition).

Let $f:[a,b]\to \mathbb{R}$. For any $x\in [a,b]$, define the quantity

$$f^{\prime }(x)\equiv \underset{t\to x}{\lim }\frac{f(t)-f(x)}{t-x}$$

provided the limit exists, in which case we say $f$ is differentiable at $x$.

Note:

• We are justified in attempting to take the limit for all $x\in [a,b]$ since they are all limit points of $[a,b]$.
• We can make same definition for $f^{\prime }(x)$ when $f$ is defined on $(a,b)$.
• Let $E=\{x|f^{\prime }(x)\text{ exists}\}$. Then, $f^{\prime }:E\to R$. We say $f$ is differentiable on $E$.

If $f^{\prime }(x)$ exists, then define $u(t)$ by :

$$u(t)\equiv \{\begin{array}{ll}\frac{f(t)-f(x)}{t-x}-f^{\prime }(x)& t\ne x\\ 0& t=x\end{array}$$

$u(t)$ satisfies

$$f(t)-f(x)=(t-x)[f^{\prime }(x)+u(t)],$$

and as $t\to 0$, $u(t)\to 0$.

Differentiability implies continuity

Rudin, 5.2

Theorem 2(Theorem).

$f^{\prime }(x)$ exists $⟹$ $f$ is continuous at $x$.

Proof

$$\underset{t\to x}{\lim }(f(t)-f(x))=\underset{t\to x}{\lim }\frac{f(t)-f(x)}{t-x}(t-x).$$

From the algebra of limits, this is equal to

$$\underset{t\to x}{\lim }\frac{f(t)-f(x)}{t-x}\underset{t\to x}{\lim }(t-x)=f^{\prime }(x)\cdot 0=0$$

Thus, ${\lim }_{t\to x}f(t)=f(x)$. ❏

The converse is NOT true!

Algebra of derivatives

Rudin, 5.3

Theorem 3(Theorem).

Suppose $f$ and $g$ are defined on $[a,b]$ and are differentiable at a point $x\in [a,b]$. Then, $f+g$, $fg$, and $f/g$ are differentiable are $x$, and

1. $(f+g{)}^{\prime }(x)=f^{\prime }(x)+g^{\prime }(x)$;
2. $(fg{)}^{\prime }(x)=f^{\prime }(x)g(x)+f(x)g^{\prime }(x)$;
3. ${\left(\frac{f}{g}\right)}^{\prime }(x)=\frac{f^{\prime }(x)g(x)-f(x)g^{\prime }(x)}{g^2(x)}$, if $g(x)\ne 0$.

Proof of 1
Follows from the algebra of limits.

Proof of 2
$f(t)g(t)-f(x)g(x)=(f(t)-f(x))g(t)+(g(t)-g(x))f(x)$.

Proof of 3
$\frac{f(t)}{g(t)}-\frac{f(x)}{g(x)}=\frac{(f(t)-f(x))g(x)-(g(t)-g(x))f(x)}{g(x)g(t)}$.

Chain rule

Rudin, 5.5

Theorem 4(Theorem).

Suppose $f$ is defined $[a,b]$, $f^{\prime }(x)$ exists at some $x\in [a,b]$, $g$ is defined on an interval $I$ which contains the range of $f$, and $g$ is differentiable at the point $f(x)$.
If $h(t)=g(f(t)),a\le t\le b$, then $h$ is differentiable at $x$, and

$$h^{\prime }(x)=g^{\prime }(f(x))f^{\prime }(x).$$

Proof
Let $y=f(x)$. By the definition of the derivative, we have

$$\begin{array}{rl}f(t)-f(x)& =(t-x)[f^{\prime }(x)+u(t)],\\ g(s)-g(y)& =(s-y)[g^{\prime }(y)+v(s)],\end{array}$$

where $t\in [a,b]$, $s\in I$, and $u(t)\to 0$ as $t\to x$, $v(s)\to 0$ as $s\to y$. Now,

$$\begin{array}{rl}h(t)-h(x)=& g(f(t))-g(f(x))\\ =& (f(t)-f(x))(g^{\prime }(f(x))+v(f(t)))\\ =& (t-x)(f^{\prime }(x)+u(t))(g^{\prime }(f(x))+v(f(t)))\\ & \\ ⟹\frac{h(t)-h(x)}{t-x}=& (f^{\prime }(x)+u(t))(g^{\prime }(f(x))+v(f(t)))\\ ⟹h^{\prime }(x)=& \underset{t\to x}{\lim }(f^{\prime }(x)+u(t))(g^{\prime }(f(x))+v(f(t)))\end{array}$$

We know $u(t)$ vanishes as $t\to x$. As $t\to x$, $f(t)\to f(x)$ since $f$ is continuous at $x$. As $f(t)\to f(x)$, $v(f(t))\to 0$ . Thus,

$$h^{\prime }(x)=f^{\prime }(x)g^{\prime }(f(x)).$$

❏