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Home Calculus Chapter 4Derivative Definition

CALC-4.1

4-5 hours

Foundation

Derivative Definition and Basic Rules

Master the rigorous limit definition of derivatives, understand one-sided derivatives and their relationship to differentiability, prove the fundamental connection between differentiability and continuity, and learn essential algebraic rules for computing derivatives efficiently.

Sections

Theorems

15+

Examples

Questions

Learning Objectives

By the end of this module, you will be able to:

Define the derivative of a function at a point using the limit of the difference quotient

Interpret the derivative geometrically as the slope of the tangent line

Understand and compute one-sided (left and right) derivatives

State and prove the theorem that differentiability implies continuity

Recognize functions that are continuous but not differentiable at certain points

Apply the basic derivative formulas for polynomial, exponential, and trigonometric functions

Derive and apply the sum, difference, product, and quotient rules

Use multiple differentiation rules together to find derivatives of complex expressions

Determine differentiability of piecewise functions at junction points

Understand the notation systems for derivatives (Leibniz, Lagrange, Newton)

Prerequisites

Before studying derivatives, ensure you have mastered:

Limit definitions and computation techniques (Chapter 2 & 3)
Continuity and its relationship to limits
Properties of elementary functions (exponential, logarithmic, trigonometric)

1. The Definition of Derivative

Understanding the limit definition and its geometric meaning

The derivative is one of the two fundamental concepts of calculus (the other being the integral). It measures the instantaneous rate of change of a function at a point, generalizing the notion of velocity from physics to arbitrary functions. The key insight is that we can compute instantaneous rates by taking limits of average rates over smaller and smaller intervals.

Historically, Newton developed derivatives (which he called "fluxions") to describe motion and physics, while Leibniz independently developed the same concepts with notation emphasizing infinitesimals. The modern rigorous definition using limits was formalized by Cauchy and Weierstrass in the 19th century.

In this section, we begin with the precise ε-δ style definition of the derivative, then explore its various equivalent formulations and geometric interpretations.

Definition 4.1: Derivative at a Point

Let $f$ be a function defined on an open interval containing $x_0$ (i.e., on some neighborhood $U(x_0)$ ). 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 $x_0$ .

Common notations for the derivative include:

f'(x_0), \quad \frac{df}{dx}\bigg|_{x=x_0}, \quad \frac{dy}{dx}\bigg|_{x=x_0}, \quad y'(x_0), \quad Df(x_0)

If this limit does not exist (or is infinite), we say $f$ is not differentiable at $x_0$ .

Remark 4.1: Equivalent Formulations

By making the substitution $x = x_0 + h$ (or equivalently $x = x_0 + \Delta x$ ), the derivative definition can be written in several equivalent forms:

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

f'(x_0) = \lim_{\Delta x \to 0} \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}

The expression $\frac{f(x_0 + h) - f(x_0)}{h}$ is called the difference quotient. It represents the average rate of change of $f$ over the interval $[x_0, x_0 + h]$ .

Geometric interpretation: The difference quotient is the slope of the secant linethrough points $(x_0, f(x_0))$ and $(x_0 + h, f(x_0 + h))$ . As $h \to 0$ , this secant line approaches the tangent line, and its slope approaches $f'(x_0)$ .

Physical interpretation: If $f(t)$ represents position at time $t$ , then $f'(t_0)$ is the instantaneous velocity at time $t_0$ .

Definition 4.2: Derivative Function

If $f(x)$ is differentiable at every point of an open interval $(a, b)$ , then $f'(x)$ defines a new function on $(a, b)$ , called the derivative function (or simply the derivative) of $f$ :

f': (a,b) \to \mathbb{R}, \quad x \mapsto f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

We write $f \in D(a, b)$ or say " $f$ is differentiable on $(a, b)$ " to denote that $f$ has a derivative at every point of the interval.

Example 4.1: Derivative from Definition: Power Function

Problem: Use the definition to find the derivative of $f(x) = x^2$ .

Solution:

Using the definition with $f(x) = x^2$ :

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

Expanding the numerator:

= \lim_{h \to 0} \frac{x_0^2 + 2x_0 h + h^2 - x_0^2}{h} = \lim_{h \to 0} \frac{2x_0 h + h^2}{h}

Factoring out $h$ and canceling:

= \lim_{h \to 0} \frac{h(2x_0 + h)}{h} = \lim_{h \to 0} (2x_0 + h) = 2x_0

Conclusion: $(x^2)' = 2x$ , which matches the power rule with $n = 2$ .

Example 4.2: Derivative from Definition: Exponential Function

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

Solution:

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

Using the exponential property $e^{x+h} = e^x \cdot e^h$ :

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

Recall the fundamental limit $\lim_{h \to 0} \frac{e^h - 1}{h} = 1$ :

f'(x) = e^x \cdot 1 = e^x

Remarkable property: The function $e^x$ is its own derivative! This is why $e$ is the "natural" base for exponentials.

Example 4.3: Derivative from Definition: Sine Function

Problem: Prove that $(\sin x)' = \cos x$ using the definition.

Solution:

(\sin x)' = \lim_{h \to 0} \frac{\sin(x+h) - \sin x}{h}

Using the angle addition formula $\sin(x+h) = \sin x \cos h + \cos x \sin h$ :

= \lim_{h \to 0} \frac{\sin x \cos h + \cos x \sin h - \sin x}{h}

Rearranging:

= \lim_{h \to 0} \left[ \sin x \cdot \frac{\cos h - 1}{h} + \cos x \cdot \frac{\sin h}{h} \right]

Using $\lim_{h \to 0} \frac{\sin h}{h} = 1$ and $\lim_{h \to 0} \frac{\cos h - 1}{h} = 0$ :

= \sin x \cdot 0 + \cos x \cdot 1 = \cos x

Example 4.4: Derivative from Definition: Cosine Function

Problem: Prove that $(\cos x)' = -\sin x$ .

Solution:

(\cos x)' = \lim_{h \to 0} \frac{\cos(x+h) - \cos x}{h}

Using $\cos(x+h) = \cos x \cos h - \sin x \sin h$ :

= \lim_{h \to 0} \frac{\cos x \cos h - \sin x \sin h - \cos x}{h}

= \lim_{h \to 0} \left[ \cos x \cdot \frac{\cos h - 1}{h} - \sin x \cdot \frac{\sin h}{h} \right]

= \cos x \cdot 0 - \sin x \cdot 1 = -\sin x

Example 4.5: Derivative from Definition: Natural Logarithm

Problem: Prove that $(\ln x)' = \frac{1}{x}$ for $x > 0$ .

Solution:

(\ln x)' = \lim_{h \to 0} \frac{\ln(x+h) - \ln x}{h} = \lim_{h \to 0} \frac{1}{h} \ln\frac{x+h}{x}

= \lim_{h \to 0} \frac{1}{h} \ln\left(1 + \frac{h}{x}\right) = \lim_{h \to 0} \frac{1}{x} \cdot \frac{x}{h} \ln\left(1 + \frac{h}{x}\right)

Let $u = h/x$ , so as $h \to 0$ , $u \to 0$ :

= \frac{1}{x} \lim_{u \to 0} \frac{\ln(1+u)}{u} = \frac{1}{x} \cdot 1 = \frac{1}{x}

Remark 4.0: Geometric Interpretation of Derivative

The derivative has a beautiful geometric interpretation that helps us visualize what differentiation means:

Secant Line

A line through two points $(x_0, f(x_0))$ and $(x_0+h, f(x_0+h))$ on the curve.

Slope = $\frac{f(x_0+h) - f(x_0)}{h}$ (difference quotient)

Tangent Line

The limiting position of secant lines as the second point approaches the first.

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

Tangent Line Equation

At point $(x_0, f(x_0))$ , the tangent line is: $y - f(x_0) = f'(x_0)(x - x_0)$

Example 4.5b: Finding Tangent Line Equations

Problem: Find the equation of the tangent line to $y = x^2$ at $x = 2$ .

Solution:

Step 1: Find the point on the curve.

At $x_0 = 2$ : $f(2) = 2^2 = 4$ , so the point is $(2, 4)$ .

Step 2: Find the derivative (slope of tangent).

$f'(x) = 2x$ , so $f'(2) = 4$

Step 3: Write the tangent line equation.

y - 4 = 4(x - 2) \implies y = 4x - 4

Example 4.5c: When Tangent Line is Horizontal

Problem: Find points where the tangent to $y = x^3 - 3x$ is horizontal.

Solution:

A horizontal tangent means slope = 0, i.e., $f'(x) = 0$ .

f'(x) = 3x^2 - 3 = 0 \implies x^2 = 1 \implies x = \pm 1

The points are:

At $x = 1$ : $y = 1 - 3 = -2$ , point $(1, -2)$
At $x = -1$ : $y = -1 + 3 = 2$ , point $(-1, 2)$

2. One-Sided Derivatives

Left and right derivatives and characterization of differentiability

Just as we defined one-sided limits, we can define one-sided derivatives by approaching the point from only one direction. These are useful for analyzing functions at endpoints, at corners, or at points where behavior differs on different sides. Understanding one-sided derivatives gives us a precise characterization of when the (two-sided) derivative exists.

Definition 4.3: One-Sided Derivatives

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

Right Derivative (derivative from the right):

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

Left Derivative (derivative from the left):

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

Theorem 4.1: 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)

In this case, $f'(x_0) = f'_+(x_0) = f'_-(x_0)$ .

Proof of Theorem 4.1:

(⇒) Suppose $f'(x_0)$ exists. Since the two-sided limit exists, both one-sided limits must exist and equal the same value.

(⇐) Suppose $f'_+(x_0) = f'_-(x_0) = L$ . Then both one-sided limits of the difference quotient equal $L$ , so the two-sided limit exists and equals $L$ .

∎

Example 4.6: The Absolute Value Function

Problem: Analyze the differentiability of $f(x) = |x|$ at $x = 0$ .

Solution:

Recall $|x| = x$ for $x \geq 0$ and $|x| = -x$ for $x < 0$ .

Right derivative:

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

Left derivative:

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

Conclusion: Since $f'_+(0) = 1 \neq -1 = f'_-(0)$ , $|x|$ is not differentiable at 0.

Example 4.7: Differentiability of x|x|

Problem: Is $f(x) = x|x|$ differentiable at $x = 0$ ?

Solution:

Note: $x|x| = x^2$ for $x \geq 0$ and $x|x| = -x^2$ for $x < 0$ .

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

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

Conclusion: Since $f'_+(0) = f'_-(0) = 0$ , $x|x|$ is differentiable at 0 with $f'(0) = 0$ .

Example 4.8: Piecewise Function Analysis

Problem: Is $f(x) = \begin{cases} x^2 & x \leq 1 \\ 2x - 1 & x > 1 \end{cases}$ differentiable at $x = 1$ ?

Solution:

First check continuity: $f(1) = 1$ , $\lim_{x \to 1^-} x^2 = 1$ , $\lim_{x \to 1^+} (2x-1) = 1$ . ✓ Continuous.

f'_-(1) = \lim_{h \to 0^-} \frac{(1+h)^2 - 1}{h} = \lim_{h \to 0^-} \frac{2h + h^2}{h} = 2

f'_+(1) = \lim_{h \to 0^+} \frac{2(1+h) - 1 - 1}{h} = \lim_{h \to 0^+} \frac{2h}{h} = 2

Conclusion: $f'_-(1) = f'_+(1) = 2$ , so $f$ is differentiable at 1 with $f'(1) = 2$ .

Example 4.8b: Piecewise Function - Not Differentiable

Problem: Is $g(x) = \begin{cases} x^2 & x \leq 1 \\ x & x > 1 \end{cases}$ differentiable at $x = 1$ ?

Solution:

Check continuity: $g(1) = 1$ , $\lim_{x \to 1^-} x^2 = 1$ , $\lim_{x \to 1^+} x = 1$ . ✓ Continuous.

g'_-(1) = \lim_{h \to 0^-} \frac{(1+h)^2 - 1}{h} = 2

g'_+(1) = \lim_{h \to 0^+} \frac{(1+h) - 1}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1

Conclusion: $g'_-(1) = 2 \neq 1 = g'_+(1)$ , so $g$ is NOT differentiable at 1.

Geometric insight: The graph has a "corner" at $x = 1$ where the slope changes abruptly from 2 to 1.

Remark 4.1b: Endpoints of Intervals

At endpoints of the domain, only one-sided derivatives can be defined:

At left endpoint $a$ of $[a, b]$ : only $f'_+(a)$ makes sense
At right endpoint $b$ of $[a, b]$ : only $f'_-(b)$ makes sense

Example: For $f(x) = \sqrt{x}$ on $[0, \infty)$ , at $x = 0$ :

f'_+(0) = \lim_{h \to 0^+} \frac{\sqrt{h} - 0}{h} = \lim_{h \to 0^+} \frac{1}{\sqrt{h}} = +\infty

So $\sqrt{x}$ has a vertical tangent at $x = 0$ .

Example 4.8c: Vertical Tangent

Problem: Analyze the differentiability of $f(x) = \sqrt[3]{x} = x^{1/3}$ at $x = 0$ .

Solution:

f'(0) = \lim_{h \to 0} \frac{\sqrt[3]{h} - 0}{h} = \lim_{h \to 0} \frac{h^{1/3}}{h} = \lim_{h \to 0} h^{-2/3} = \lim_{h \to 0} \frac{1}{h^{2/3}}

As $h \to 0$ , $h^{2/3} \to 0^+$ , so the limit is $+\infty$ .

Conclusion: $f$ is not differentiable at 0 (in the usual sense), but has a vertical tangent there. The function is continuous at 0 but the tangent line is vertical.

3. Differentiability and Continuity

The fundamental relationship between these conditions

Theorem 4.2: 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

Important: The converse is FALSE. Continuity does NOT imply differentiability.

Proof of Theorem 4.2:

Suppose $f'(x_0)$ exists. We need to show $\lim_{x \to x_0} f(x) = f(x_0)$ .

For $x \neq x_0$ , we can write:

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

Therefore $\lim_{x \to x_0} f(x) = f(x_0)$ , which is the definition of continuity at $x_0$ . ∎

∎

Corollary 4.1: Contrapositive Form

If $f$ is discontinuous at $x_0$ , then $f$ is not differentiable at $x_0$ .

Example 4.9: 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: $f'_+(0) = 1 \neq -1 = f'_-(0)$

Geometric interpretation: The graph has a "corner" at $x = 0$ where no unique tangent line exists. The left tangent has slope -1, the right tangent has slope 1.

Example 4.10: x·sin(1/x) Analysis

Problem: Analyze $f(x) = x\sin(1/x)$ for $x \neq 0$ , $f(0) = 0$ .

Solution:

Continuity at 0:

Since $|x\sin(1/x)| \leq |x| \to 0$ , by squeeze theorem $\lim_{x \to 0} f(x) = 0 = f(0)$ . ✓

Differentiability at 0:

f'(0) = \lim_{h \to 0} \frac{h\sin(1/h) - 0}{h} = \lim_{h \to 0} \sin\frac{1}{h}

This limit does not exist because $\sin(1/h)$ oscillates. So $f$ is continuous but not differentiable at 0.

Example 4.11: x²·sin(1/x) Analysis

Problem: Analyze $g(x) = x^2\sin(1/x)$ for $x \neq 0$ , $g(0) = 0$ .

Solution:

g'(0) = \lim_{h \to 0} \frac{h^2\sin(1/h) - 0}{h} = \lim_{h \to 0} h\sin\frac{1}{h}

Since $|h\sin(1/h)| \leq |h| \to 0$ , by squeeze theorem:

$g'(0) = 0$ . The extra factor of $x$ provides enough decay!

Remark 4.1c: Summary: Differentiability vs Continuity

Condition	Continuous?	Differentiable?	Example
Smooth function	✓ Yes	✓ Yes	$x^2, e^x, \sin x$
Corner point	✓ Yes	✗ No	$\|x\|$ at 0
Vertical tangent	✓ Yes	✗ No	$\sqrt[3]{x}$ at 0
Cusp	✓ Yes	✗ No	$x^{2/3}$ at 0
Jump discontinuity	✗ No	✗ No	Step function
Oscillating	✓ Yes*	✗ No	$x\sin(1/x)$ at 0

*Requires proper definition at the point (e.g., $f(0) = 0$ )

Example 4.11b: Cusp Analysis

Problem: Analyze $f(x) = x^{2/3}$ at $x = 0$ .

Solution:

Check continuity: $\lim_{x \to 0} x^{2/3} = 0 = f(0)$ . ✓

Check differentiability:

f'_+(0) = \lim_{h \to 0^+} \frac{h^{2/3} - 0}{h} = \lim_{h \to 0^+} h^{-1/3} = +\infty

f'_-(0) = \lim_{h \to 0^-} \frac{|h|^{2/3} - 0}{h} = \lim_{h \to 0^-} \frac{(-h)^{2/3}}{h} = \lim_{h \to 0^-} -|h|^{-1/3} = -\infty

Conclusion: $f$ is continuous at 0 but NOT differentiable. It has a cusp: left tangent is vertical downward, right tangent is vertical upward.

Theorem 4.2b: Local Linear Approximation

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

f(x) \approx f(x_0) + f'(x_0)(x - x_0)

This is the linear approximation or tangent line approximation. The error goes to zero faster than $(x - x_0)$ :

f(x) = f(x_0) + f'(x_0)(x - x_0) + o(x - x_0) \text{ as } x \to x_0

Application: Use to approximate $\sqrt{4.1} \approx \sqrt{4} + \frac{1}{2\sqrt{4}}(0.1) = 2.025$

4. Basic Derivative Formulas

Essential derivatives of elementary functions

Elementary Function Derivatives

(c)'

0

Constant function

(x^n)'

nx^{n-1}

Power rule (n ∈ ℝ)

(e^x)'

e^x

Natural exponential

(a^x)'

a^x \ln a

General exponential (a > 0)

(\ln x)'

\frac{1}{x}

Natural logarithm

(\log_a x)'

\frac{1}{x \ln a}

General logarithm

(\sin x)'

\cos x

Sine function

(\cos x)'

-\sin x

Cosine function

(\tan x)'

\sec^2 x

Tangent function

(\cot x)'

-\csc^2 x

Cotangent function

(\sec x)'

\sec x \tan x

Secant function

(\csc x)'

-\csc x \cot x

Cosecant function

Inverse Trigonometric

(\arcsin x)'

\frac{1}{\sqrt{1-x^2}}

|x| < 1

(\arccos x)'

-\frac{1}{\sqrt{1-x^2}}

|x| < 1

(\arctan x)'

\frac{1}{1+x^2}

x ∈ ℝ

(\text{arccot } x)'

-\frac{1}{1+x^2}

x ∈ ℝ

Example 4.12: Proving the Power Rule

Method 1: Using Binomial Theorem (n ∈ ℕ)

(x^n)' = \lim_{h \to 0} \frac{(x+h)^n - x^n}{h}

Using binomial theorem:

(x+h)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} h^k = x^n + nx^{n-1}h + \binom{n}{2}x^{n-2}h^2 + \cdots

= \lim_{h \to 0} \frac{nx^{n-1}h + O(h^2)}{h} = \lim_{h \to 0} \left[nx^{n-1} + O(h)\right] = nx^{n-1}

Example 4.13: Derivative of Square Root

Problem: Prove $(\sqrt{x})' = \frac{1}{2\sqrt{x}}$ for $x > 0$ .

Solution:

(\sqrt{x})' = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}

Rationalize by multiplying by conjugate:

= \lim_{h \to 0} \frac{(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x})}{h(\sqrt{x+h} + \sqrt{x})}

= \lim_{h \to 0} \frac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})} = \lim_{h \to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})}

= \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}} = \frac{1}{2\sqrt{x}}

Note: This confirms the power rule with $n = 1/2$ : $(x^{1/2})' = \frac{1}{2}x^{-1/2}$ .

Example 4.14: Derivative of 1/x

Problem: Prove $(1/x)' = -1/x^2$ for $x \neq 0$ .

Solution:

\left(\frac{1}{x}\right)' = \lim_{h \to 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h}

Combining fractions:

= \lim_{h \to 0} \frac{1}{h} \cdot \frac{x - (x+h)}{x(x+h)} = \lim_{h \to 0} \frac{-h}{hx(x+h)}

= \lim_{h \to 0} \frac{-1}{x(x+h)} = \frac{-1}{x \cdot x} = -\frac{1}{x^2}

Remark 4.2: Extension of Power Rule

The power rule $(x^n)' = nx^{n-1}$ holds for all real exponents $n \in \mathbb{R}$ , not just positive integers:

Negative integers: $(x^{-n})' = -nx^{-n-1}$ (can verify using quotient rule)
Fractions: $(x^{p/q})' = \frac{p}{q}x^{p/q - 1}$ (requires implicit differentiation)
Irrationals: $(x^\pi)' = \pi x^{\pi - 1}$ (requires logarithmic differentiation)

The general proof uses logarithmic differentiation: $y = x^n \Rightarrow \ln y = n\ln x$ .

Remark 4.3: Important Limits for Derivatives

These fundamental limits are essential for computing derivatives from definition:

\lim_{h \to 0} \frac{\sin h}{h} = 1

\lim_{h \to 0} \frac{\cos h - 1}{h} = 0

\lim_{h \to 0} \frac{e^h - 1}{h} = 1

\lim_{h \to 0} \frac{\ln(1+h)}{h} = 1

\lim_{h \to 0} \frac{a^h - 1}{h} = \ln a

\lim_{h \to 0} \frac{(1+h)^n - 1}{h} = n

Example 4.14b: Derivative of General Exponential

Problem: Prove $(a^x)' = a^x \ln a$ for $a > 0, a \neq 1$ .

Solution:

Write $a^x = e^{x \ln a}$ using the identity $a = e^{\ln a}$ .

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

= a^x \lim_{h \to 0} \frac{a^h - 1}{h}

Using the fundamental limit $\lim_{h \to 0} \frac{a^h - 1}{h} = \ln a$ :

(a^x)' = a^x \ln a

Special case: When $a = e$ , $\ln e = 1$ , so $(e^x)' = e^x$ .

Example 4.14c: Derivative of General Logarithm

Problem: Prove $(\log_a x)' = \frac{1}{x \ln a}$ for $a > 0, a \neq 1, x > 0$ .

Solution:

Use the change of base formula: $\log_a x = \frac{\ln x}{\ln a}$

(\log_a x)' = \left(\frac{\ln x}{\ln a}\right)' = \frac{1}{\ln a} \cdot (\ln x)' = \frac{1}{\ln a} \cdot \frac{1}{x} = \frac{1}{x \ln a}

Special case: When $a = e$ , $\log_e x = \ln x$ , so $(\ln x)' = \frac{1}{x}$ .

Remark 4.3b: Hyperbolic Functions Derivatives

The hyperbolic functions and their derivatives mirror trigonometric functions:

(\sinh x)' = \cosh x

where $\sinh x = \frac{e^x - e^{-x}}{2}$

(\cosh x)' = \sinh x

where $\cosh x = \frac{e^x + e^{-x}}{2}$

(\tanh x)' = \text{sech}^2 x

Similar to $(\tan x)' = \sec^2 x$

(\text{coth } x)' = -\text{csch}^2 x

Similar to $(\cot x)' = -\csc^2 x$

Note: Unlike trig functions, $(\cosh x)' = \sinh x$ (positive), not $-\sinh x$ .

5. Differentiation Rules

Product, quotient, and sum rules

Theorem 4.3: Algebraic Rules

Sum Rule:

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

Product Rule:

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

Quotient Rule:

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

Proof of Theorem 4.3 (Quotient Rule):

We derive the quotient rule from the product rule. Write $f/g = f \cdot g^{-1}$ .

First, we need $(g^{-1})' = -g'/g^2$ (which can be proven using the definition).

\left(\frac{f}{g}\right)' = (f \cdot g^{-1})' = f' \cdot g^{-1} + f \cdot (g^{-1})'

= \frac{f'}{g} + f \cdot \left(-\frac{g'}{g^2}\right) = \frac{f'}{g} - \frac{fg'}{g^2}

= \frac{f'g - fg'}{g^2}

∎

Example 4.15: Product Rule Applications

(a) $(x^2 e^x)'$

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

(b) $(x\sin x)'$

= (x)' \cdot \sin x + x \cdot (\sin x)' = \sin x + x\cos x

= e^x \cos x + e^x(-\sin x) = e^x(\cos x - \sin x)

(d) $(x^3 \ln x)'$ for $x > 0$

= 3x^2 \ln x + x^3 \cdot \frac{1}{x} = 3x^2 \ln x + x^2 = x^2(3\ln x + 1)

Example 4.16: Quotient Rule Applications

(a) $(\tan x)' = \left(\frac{\sin x}{\cos x}\right)'$

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

(b) $(\cot x)' = \left(\frac{\cos x}{\sin x}\right)'$

= \frac{-\sin x \cdot \sin x - \cos x \cdot \cos x}{\sin^2 x} = \frac{-(\sin^2 x + \cos^2 x)}{\sin^2 x} = -\frac{1}{\sin^2 x} = -\csc^2 x

= \frac{1 \cdot (1+x^2) - x \cdot 2x}{(1+x^2)^2} = \frac{1+x^2 - 2x^2}{(1+x^2)^2} = \frac{1-x^2}{(1+x^2)^2}

(d) $\left(\frac{e^x}{x^2}\right)'$

= \frac{e^x \cdot x^2 - e^x \cdot 2x}{x^4} = \frac{e^x(x^2 - 2x)}{x^4} = \frac{e^x(x-2)}{x^3}

Example 4.17: Combined Rules - Complex Examples

(a) Find $\left(\frac{x^2 \sin x}{e^x}\right)'$

Let $u = x^2 \sin x$ , then $u' = 2x\sin x + x^2\cos x$ .

\left(\frac{u}{e^x}\right)' = \frac{u' \cdot e^x - u \cdot e^x}{e^{2x}} = \frac{(2x\sin x + x^2\cos x - x^2\sin x)e^x}{e^{2x}}

= \frac{2x\sin x + x^2(\cos x - \sin x)}{e^x}

(b) Find $\left(\frac{\ln x}{x^2}\right)'$ for $x > 0$

= \frac{\frac{1}{x} \cdot x^2 - \ln x \cdot 2x}{x^4} = \frac{x - 2x\ln x}{x^4} = \frac{1 - 2\ln x}{x^3}

Using product rule on three functions: $(uvw)' = u'vw + uv'w + uvw'$

= 2x \cdot e^x \sin x + x^2 \cdot e^x \sin x + x^2 e^x \cos x

= x e^x (2\sin x + x\sin x + x\cos x) = xe^x[(x+2)\sin x + x\cos x]

Remark 4.4: Derivative Notation Comparison

Notation	Name	Best For
$f'(x), y'$	Lagrange	Compact notation, function emphasis
$\frac{dy}{dx}, \frac{df}{dx}$	Leibniz	Chain rule, related rates, integration
$\dot{y}, \dot{x}$	Newton	Time derivatives (physics)
$Df, D_x f$	Operator	Abstract analysis, operators

Remark 4.5: Common Differentiation Mistakes

❌ Wrong: $(fg)' = f'g'$

✓ Correct: $(fg)' = f'g + fg'$ (Product rule)

❌ Wrong: $\left(\frac{f}{g}\right)' = \frac{f'}{g'}$

✓ Correct: $\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$

❌ Wrong: $(x^n)' = x^{n-1}$

✓ Correct: $(x^n)' = nx^{n-1}$ (Don't forget the coefficient!)

Practice Quiz

Derivative Definition Quiz

Questions

Correct

Accuracy

What is the formal definition of the derivative

f'(x_0)

Easy

Not attempted

For

f(x) = |x|

, what can we say about

f'(0)

Medium

Not attempted

Which statement about differentiability and continuity is correct?

Easy

Not attempted

Calculate

(x^3 \cdot e^x)'

using the product rule.

Medium

Not attempted

What is

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

for

x \neq 0

Medium

Not attempted

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

, what is

f'(x_0)

Easy

Not attempted

For

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

(

x \neq 0

f(0) = 0

, is

f

differentiable at 0?

Hard

Not attempted

What is

(\tan x)'

Easy

Not attempted

For

f(x) = \sqrt{x}

, what is

f'(x)

for

x > 0

Easy

Not attempted

The quotient rule states

(f/g)' =

Medium

Not attempted

Find

(x^x)'

for

x > 0

Hard

Not attempted

What is the geometric meaning of

f'(x_0)

Easy

Not attempted

What is

(\ln x)'

for

x > 0

Easy

Not attempted

Find

(e^x \sin x)'

Medium

Not attempted

What is

(\sec x)'

Medium

Not attempted

f(x) = x^3 - 3x

, find

f'(2)

Easy

Not attempted

What is

\frac{d}{dx}(\arctan x)

Medium

Not attempted

f

is differentiable at

a

, which is guaranteed?

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.

When should I use the definition vs differentiation rules?

Use the definition for: proving rules, piecewise functions at junction points, functions defined by limits/series, checking existence. Use rules for efficient computation.

What are the most common mistakes in differentiation?

Common mistakes include: (1) Forgetting to apply product rule when multiplying functions, (2) Mixing up signs in quotient rule, (3) Forgetting the coefficient in power rule (writing x^(n-1) instead of nx^(n-1)), (4) Not checking domain restrictions.

How do I find the equation of a tangent line?

At point (x₀, f(x₀)): (1) Find f'(x₀) for the slope, (2) Use point-slope form: y - f(x₀) = f'(x₀)(x - x₀). This can be rewritten as y = f(x₀) + f'(x₀)(x - x₀).

What does it mean if the derivative is zero at a point?

If f'(x₀) = 0, the tangent line is horizontal at x₀. This point is called a critical point. It could be a local maximum, local minimum, or inflection point - further analysis (second derivative test or first derivative test) is needed.

Can a function be continuous everywhere but differentiable nowhere?

Yes! The Weierstrass function is continuous everywhere but differentiable nowhere. It's defined as an infinite sum of cosines with rapidly increasing frequencies. This counterintuitive example shows continuity and differentiability are very different properties.

What is logarithmic differentiation and when should I use it?

Logarithmic differentiation involves taking ln of both sides, then differentiating. Use it for: (1) Functions like x^x where variable is in both base and exponent, (2) Products of many functions, (3) Complicated quotients. The technique simplifies multiplication to addition.

How do I check if a piecewise function is differentiable at the junction?

At junction point x₀: (1) Check continuity: lim(x→x₀⁻) f(x) = lim(x→x₀⁺) f(x) = f(x₀), (2) Check derivatives match: f'₋(x₀) = f'₊(x₀). Both conditions must hold for differentiability.

Key Takeaways

Definition

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

Key Relationship

Differentiable ⟹ Continuous (not vice versa)

One-Sided Test

$f'$ exists ⟺ $f'_+ = f'_-$

Product Rule

(fg)' = f'g + fg'

Quotient Rule

(f/g)' = \frac{f'g - fg'}{g^2}

Power Rule

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

Exponential

(e^x)' = e^x, \quad (a^x)' = a^x \ln a

Logarithm

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

Trigonometric

(\sin x)' = \cos x, \, (\cos x)' = -\sin x

Tangent Line

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

What's Next?

In the next module, you will learn about the Chain Rule for differentiating composite functions, and Higher-Order Derivatives including the second derivative and beyond.

Chain rule: $(f \circ g)'(x) = f'(g(x)) \cdot g'(x)$
Implicit differentiation for equations like $x^2 + y^2 = 1$
Higher derivatives: $f'', f''', f^{(n)}$
Leibniz formula for nth derivative of products
Parametric differentiation for curves $(x(t), y(t))$
Inverse function derivatives

Study Tips

Practice Strategy

Start with simple functions, then gradually combine rules. Always verify your answers using the definition for a few cases.

Memorization

Memorize the basic formulas table. Create flashcards for quick review. Understanding the proofs helps remember the formulas.

Common Pitfalls

Don't forget the coefficient in power rule. Always check if product/quotient rule is needed before differentiating.

Verification

Check your answers by substituting specific values. Use graphing tools to visualize tangent lines.