Math4121 Lecture 1

Chapter 5: Differentiation

The derivative of a real function

Definition 5.1

Let $f$ be a real-valued function on an interval $[a,b]$ ($f: [a,b] \to \mathbb{R}$).

We say that $f$ is differentiable at a point $x\in [a,b]$ if the limit

exists.

Then we defined the derivative of $f$, $f'$, a function whose domain is the set of all $x\in [a,b]$ at which $f$ is differentiable, by

Theorem 5.2

Let $f:[a,b]\to \mathbb{R}$. If $f$ is differentiable at $x\in [a,b]$, then $f$ is continuous at $x$.

Proof:

Recall Definition 4.5 

$f$ is continuous at $x$ if $\forall \epsilon > 0, \exists \delta > 0$ such that if $|t-x| < \delta$, then $|f(t)-f(x)| < \epsilon$.

Whenever you see a limit, you should think of this definition.

We need to show that $\lim_{t\to x} f(t) = f(x)$.

Equivalently, we need to show that

So for $t\ne x$, since $f$ is differentiable at $x$, we have

Therefore, differentiable is a stronger condition than continuous.

There exists some function that is continuous but not differentiable.

For example, $f(x) = |x|$ is continuous at $x=0$, but not differentiable at $x=0$.

We can see that the left-hand limit and the right-hand limit are not the same.

$\lim_{t\to 0^-} \frac{|t|-|0|}{t-0} = -1 \quad \text{and} \quad \lim_{t\to 0^+} \frac{|t|-|0|}{t-0} = 1$

Therefore, the limit does not exist. for $f(x) = |x|$ at $x=0$.

Theorem 5.3

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

(a) $(f+g)'(x) = f'(x) + g'(x)$
(b) $(fg)'(x) = f'(x)g(x) + f(x)g'(x)$
(c) $\left(\frac{f}{g}\right)'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}$, provided $g(x)\ne 0$

Proof:

Since the limit of product is the product of the limits, we can use the definition of the derivative to prove the theorem.

(a)

(b)

Since $f$ is differentiable at $x$, we have $\lim_{t\to x} f(t) = f(x)$.

(c)