Section 1: Definition of Derivative

Definition 1.1: Derivative at a Point

Let $f$ be a function defined on an open interval containing $x_0$ . If the limit

\lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0}

exists and is finite, we say $f$ is differentiable at $x_0$ , and we call this limit the derivative of $f$ at $x_0$ , denoted $f'(x_0)$ .

Equivalent form: $f'(x_0) = \lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)}{h}$

Example 1.1: Derivative from Definition: Power Function

Prove: $(x^2)' = 2x$ using the definition.

Proof:

f'(x_0) = \lim_{h \to 0} \frac{(x_0 + h)^2 - x_0^2}{h} = \lim_{h \to 0} \frac{2x_0 h + h^2}{h} = \lim_{h \to 0} (2x_0 + h) = 2x_0

Therefore $(x^2)' = 2x$ . ∎

Example 1.2: Derivative from Definition: Exponential

Prove: $(e^x)' = e^x$ using the definition.

Proof:

f'(x) = \lim_{h \to 0} \frac{e^{x+h} - e^x}{h} = e^x \lim_{h \to 0} \frac{e^h - 1}{h} = e^x \cdot 1 = e^x

Therefore $(e^x)' = e^x$ . ∎

Section 2: One-Sided Derivatives

Definition 2.1: One-Sided Derivatives

Let $f$ be defined on a neighborhood of $x_0$ .

Right derivative: $f'_+(x_0) = \lim_{h \to 0^+} \frac{f(x_0 + h) - f(x_0)}{h}$
Left derivative: $f'_-(x_0) = \lim_{h \to 0^-} \frac{f(x_0 + h) - f(x_0)}{h}$

Theorem 2.1: Characterization of Differentiability

A function $f$ is differentiable at $x_0$ if and only if both one-sided derivatives exist and are equal:

f'(x_0) \text{ exists} \iff f'_+(x_0) = f'_-(x_0)

Example 2.1: Absolute Value Function

Analyze: $f(x) = |x|$ at $x = 0$ .

Solution:

$f'_+(0) = \lim_{h \to 0^+} \frac{|h|}{h} = 1$

$f'_-(0) = \lim_{h \to 0^-} \frac{|h|}{h} = -1$

Since $f'_+(0) \neq f'_-(0)$ , $f$ is not differentiable at 0. ∎

Section 3: Differentiability and Continuity

Theorem 3.1: Differentiability Implies Continuity

If $f$ is differentiable at $x_0$ , then $f$ is continuous at $x_0$ :

f'(x_0) \text{ exists} \implies f \text{ is continuous at } x_0

Proof of Theorem 3.1:

Suppose $f'(x_0)$ exists. For $x \neq x_0$ :

f(x) - f(x_0) = \frac{f(x) - f(x_0)}{x - x_0} \cdot (x - x_0)

Taking the limit as $x \to x_0$ :

\lim_{x \to x_0} [f(x) - f(x_0)] = f'(x_0) \cdot 0 = 0

Therefore $\lim_{x \to x_0} f(x) = f(x_0)$ . ∎

∎

Example 3.1: Continuous but Not Differentiable

The function $f(x) = |x|$ demonstrates that continuity ≠ differentiability:

Continuous at 0: $\lim_{x \to 0} |x| = 0 = |0|$ ✓

Not differentiable at 0: $f'_+(0) = 1 \neq -1 = f'_-(0)$ ✗

Section 4: Basic Derivative Formulas

Elementary Function Derivatives

(c)' = 0

Constant

(x^n)' = nx^{n-1}

Power rule

(e^x)' = e^x

Natural exponential

(\ln x)' = \frac{1}{x}

Natural logarithm

(\sin x)' = \cos x

Sine

(\cos x)' = -\sin x

Cosine

Section 5: Differentiation Rules

Theorem 5.1: Sum Rule

(f \pm g)'(x) = f'(x) \pm g'(x)

Theorem 5.2: Product Rule

(f \cdot g)'(x) = f'(x) g(x) + f(x) g'(x)

Proof of Product Rule:

We show $(fg)'(x) = f'(x)g(x) + f(x)g'(x)$ .

(fg)'(x) = \lim_{h \to 0} \frac{f(x+h)g(x+h) - f(x)g(x)}{h}

Add and subtract $f(x+h)g(x)$ :

= \lim_{h \to 0} \frac{f(x+h)[g(x+h) - g(x)] + [f(x+h) - f(x)]g(x)}{h}

Using limit laws:

= f(x)g'(x) + f'(x)g(x)

∎

Theorem 5.3: Quotient Rule

\left(\frac{f}{g}\right)'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}

Example 5.1: Product Rule Application

Find: $(x^2 e^x)'$

(x^2 e^x)' = (x^2)' \cdot e^x + x^2 \cdot (e^x)' = 2xe^x + x^2 e^x = x(x+2)e^x

Example 5.2: Quotient Rule Application

Find: $(\tan x)'$

$\tan x = \frac{\sin x}{\cos x}$

(\tan x)' = \frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x

Section 6: Higher-Order Derivatives

Definition 6.1: n-th Derivative

The n-th derivative of $f$ is defined recursively:

f^{(n)}(x) = (f^{(n-1)})'(x)

with $f^{(0)}(x) = f(x)$ .

Example 6.1: Higher Derivatives

For $f(x) = x^4$ :

$f'(x) = 4x^3$
$f''(x) = 12x^2$
$f'''(x) = 24x$
$f^{(4)}(x) = 24$
$f^{(n)}(x) = 0$ for $n > 4$

Section 7: Geometric Interpretation

Theorem 7.1: Tangent Line

If $f$ is differentiable at $x_0$ , the equation of the tangent line is:

y = f(x_0) + f'(x_0)(x - x_0)

Example 7.1: Tangent Line

Find the tangent line to $f(x) = x^2$ at $x = 1$ .

$f(1) = 1$ , $f'(1) = 2$

Tangent line: $y = 1 + 2(x - 1) = 2x - 1$

Practice Quiz: Derivative Definition and Basic Rules

10

Questions

0

Correct

0%

Accuracy

1

What is the formal definition of the derivative

f'(x_0)

?

Easy

Not attempted

2

For

f(x) = |x|

, what can we say about

f'(0)

?

Medium

Not attempted

3

Which statement about differentiability and continuity is correct?

Easy

Not attempted

4

Calculate

(x^3 \cdot e^x)'

using the product rule.

Medium

Not attempted

5

What is

(\frac{\sin x}{x})'

for

x \neq 0

?

Medium

Not attempted

6

If

f'_+(x_0) = f'_-(x_0) = 2

, what is

f'(x_0)

?

Easy

Not attempted

7

For

f(x) = x^2 \sin(1/x)

(

x \neq 0

),

f(0) = 0

, is

f

differentiable at 0?

Hard

Not attempted

8

What is

(\tan x)'

?

Easy

Not attempted

9

For

f(x) = \sqrt{x}

, what is

f'(x)

for

x > 0

?

Easy

Not attempted

10

The quotient rule states

(f/g)' =

?

Medium

Not attempted

Frequently Asked Questions

What is the geometric interpretation of the derivative?

The derivative f'(x₀) represents the slope of the tangent line to the graph of f at (x₀, f(x₀)). It is the limit of secant line slopes as the second point approaches (x₀, f(x₀)).

Why does differentiability imply continuity but not vice versa?

If f'(x₀) exists, we can write f(x) - f(x₀) = [(f(x) - f(x₀))/(x - x₀)] · (x - x₀). As x → x₀, first factor → f'(x₀), second → 0, so f(x) → f(x₀). The converse fails because |x| has a corner at 0.

How do I remember the quotient rule?

'Low d-high minus high d-low, over low squared': (f/g)' = (g·f' - f·g')/g². Low = denominator g, high = numerator f, d = derivative.

What's the difference between f'(x₀) and f'₊(x₀)?

f'(x₀) requires limits from both sides to be equal. f'₊(x₀) only considers x → x₀⁺. The derivative exists iff f'₊(x₀) = f'₋(x₀).

Can a function have a discontinuous derivative?

Yes! f(x) = x²sin(1/x) for x ≠ 0 and f(0) = 0 is differentiable everywhere, but f'(x) is discontinuous at 0 because cos(1/x) oscillates.