The Derivative

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Section 5.1 The Derivative

Definition 5.1. Interior Point.

For a metric

X

and a set

S \subseteq X,

a point

a \in S

is called an interior point if there is some

ε > 0

such that

B_{ε} (a) \subseteq S .

In particular, if

S \subseteq R, a \in S

is an interior point iff exists

ε > 0

such that

(a - ε, a + ε) \subseteq S .

Definition 5.2. Derivative.

Suppose

S \subseteq R

and

a

is an interior point of

S_{0} .

We'll say a function

f : S \to R

is differentiable at

a

if

f^{'} (a) := lim_{x \to a} \frac{f (x) - f (a)}{x - a}

exists.

Theorem 5.3. Differentiable Implies Continuous.

If

S \subseteq R, a \in S

is an interior point, and

f : S \to R

is differentiable at

a,

then

f

is continuous at

a .

Proof.

We need to show that

lim_{x \to a} f (x) = f (a) .

If

x \in S ∖ {a}

\begin{aligned} f (x) & = \frac{f (x) - f (a)}{x - a} (x - a) + f (a) \\ ⟹ lim_{x \to a} f (x) & = f^{'} (a) \cdot 0 + f (a) = f (a) . \end{aligned}

Theorem 5.4. Derivative Rules.

If

S \subseteq R, a \in S

is an interior point,

f, g : S \to R

are differentiable at

a

and

α \in R,

then

🔗
$(f + g)^{'} (a) = f^{'} (a) + g^{'} (a);$
🔗
$(α f)^{'} (a) = α f^{'} (a);$
🔗
$(f g)^{'} (a) = f^{'} (a) g (a) + f (a) g^{'} (a);$
🔗
${(\frac{f}{g})}^{'} (a) = \frac{g (a) f^{'} (a) - f (a) g^{'} (a)}{(g (a))^{2}}$ if $g (a) \neq 0 .$

Proof.

(1), (2) are exercises

For $x \in S ∖ {a},$

$\begin{array}{r} \frac{(f g) (x) - (f g) (a)}{x - a} = \frac{f (x) g (x) - f (a) g (a)}{x - a} = (\frac{f (x) - f (a)}{x - a}) g (x) + f (a) (\frac{g (x) - g (a)}{x - a}) \end{array}$

Know

$\begin{aligned} lim_{x \to a} \frac{f (x) - f (a)}{x - a} & = f^{'} (a); \\ lim_{x \to a} \frac{g (x) - g (a)}{x - a} & = g^{'} (a) \end{aligned}$

and since $g$ is continuous at $a,$

$lim_{x \to a} g (x) = g (a) .$

Therefore

$\begin{array}{r} lim_{x \to a} \frac{(f g) (x) - (f g) (a)}{x - a} = f^{'} (a) g (a) + f (a) g^{'} (a) . \end{array}$
To make sense of $(\frac{f}{g}) (a),$ let $T = {x \in S : g (x) \neq 0},$ and we'll view $\frac{f}{g}$ as a function $\frac{f}{g} : T \to R .$ Since $g (a) \neq 0$ and $g$ is continuous at $a,$ there exists $δ > 0$ such that $g (x) \neq 0$ whenever $x \in (a - δ, a + δ) .$ So $(a - δ, a + δ) \subseteq T$ and $a$ is an interior point of $T .$ The rest of the computation is similar to the one in (3).

Corollary 5.5. Power Rule.

If

n \in Z_{+}

and

f : R \to R

is given by

f (x) = x^{n},

then

f

is differentiable on

R

and

f^{'} (x) = {\begin{cases} 0 & n = 0 \\ n x^{n - 1} & n \geq 1. \end{cases}

[Here

0^{0} = 1 .

]

Proof.

n = 0, 1

are exercises Assume the result holds for some

n \geq 1 .

Set

f (x) = x^{n + 1}, g (x) = x^{n}, h (x) = x

so

f = g h .

Then

\begin{array}{r} f^{'} (x) = g^{'} (x) h (x) + g (x) h^{'} (x) = n x^{n - 1} \cdot x + x^{n} \cdot 1 = (n + 1) x^{n} . \end{array}

Remark 5.6.

We could've defined derivatives of functions

f : R \to C

(or

f : S \to C, S \subseteq R

) in the same way. Then

f

is differentiable at at point

a

iff

Re (f)

and

Im (f)

are, in which case

\begin{aligned} (Re (f))^{'} (a) & = Re (f^{'} (a)); \\ (Im (f))^{'} (a) & = Im (f^{'} (a)) . \end{aligned}

One can also define derivatives on functions

f : C \to C

in the same way

\begin{array}{r} f^{'} (a) = lim_{x \to a} \frac{f (x) - f (a)}{x - a}, \end{array}

but the theory is much different. In some sense, not many functions on

C

are differentiable.

Theorem 5.7. Chain Rule.

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