NoteNextra-origin/content/Math4121/Math4121_L1.md

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

\lim_{t\to x} \frac{f(t)-f(x)}{t-x}

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

f'(x) = \lim_{t\to x} \frac{f(t)-f(x)}{t-x}

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

\lim_{t\to x} (f(t)-f(x)) = 0

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

\begin{aligned}
\lim_{t\to x} (f(t)-f(x)) &= \lim_{t\to x} \left(\frac{f(t)-f(x)}{t-x}\right)(t-x) \\
&= \lim_{t\to x} \left(\frac{f(t)-f(x)}{t-x}\right) \lim_{t\to x} (t-x) \\
&= f'(x) \cdot 0 \\
&= 0
\end{aligned}

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)

\begin{aligned}
(f+g)'(x) &= \lim_{t\to x} \frac{(f+g)(t)-(f+g)(x)}{t-x} \\
&= \lim_{t\to x} \frac{f(t)-f(x)}{t-x} + \lim_{t\to x} \frac{g(t)-g(x)}{t-x} \\
&= f'(x) + g'(x)
\end{aligned}

(b)

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

\begin{aligned}
(fg)'(x) &= \lim_{t\to x} \left(\frac{f(t)g(t)-f(x)g(x)}{t-x}\right) \\
&= \lim_{t\to x} \left(f(t)\frac{g(t)-g(x)}{t-x} + g(x)\frac{f(t)-f(x)}{t-x}\right) \\
&= f(t) \lim_{t\to x} \frac{g(t)-g(x)}{t-x} + g(x) \lim_{t\to x} \frac{f(t)-f(x)}{t-x} \\
&= f(x)g'(x) + g(x)f'(x)
\end{aligned}

(c)

  
\begin{aligned}
\left(\frac{f}{g}\right)'(x) &= \lim_{t\to x}\left(\frac{f(t)g(x)}{g(t)g(x)} - \frac{f(x)g(x)}{g(t)g(x)}\right) \\
&= \frac{1}{g(t)g(x)}\left(\lim_{t\to x} (f(t)g(x)-f(x)g(t))\right) \\
\end{aligned}