Unit 2: Differentiation: Definition and Fundamental Properties

1. INTRODUCTION

Calculus isn't just math—it's the math of motion. While basic math looks at static snapshots (boring), Calculus zooms into the "right now." Want to know your exact speed at the moment you hit a speed trap, not just your average for the trip? That's the derivative. It's the ultimate tool for tracking change in real-time, whether you're modeling a viral TikTok trend, stock market dips, or rocket launches. Let's move from the "average" to the "instant."

2. ALL KEY CONCEPTS, TERMS, FOUNDATIONAL KNOWLEDGE, and PRINCIPLES

Foundational Terms

• Secant Line: A line passing through two distinct points on a curve, representing the average rate of change over an interval.
• Tangent Line: A line that "just touches" a curve at a single point, representing the instantaneous rate of change at that point.
• Difference Quotient: The expression $\frac{f(x+h) - f(x)}{h}$, which represents the slope of a secant line.
• Derivative: The limit of the difference quotient as the interval between two points approaches zero. It represents the slope of the tangent line or the instantaneous rate of change.
• Differentiability: A property of a function where a derivative exists at a given point.

Mathematical Notations

• Leibniz Notation: $\frac{dy}{dx}$ (read as "the derivative of $y$ with respect to $x$").
• Lagrange Notation: $f'(x)$ (read as "f prime of x").
• Operator Notation: $\frac{d}{dx}[f(x)]$ (an instruction to differentiate the function $f$).

Key Principles

• The Power Rule: A shortcut for differentiating functions of the form $x^n$.
• Linearity of the Derivative: The derivative of a sum is the sum of the derivatives, and constants can be factored out.
• Product and Quotient Rules: Specific formulas for differentiating the product or division of two functions.
• Differentiability-Continuity Relationship: A fundamental theorem stating that if a function is differentiable at a point, it must also be continuous there.

3. IN-DEPTH EXPLANATION of EVERY CONCEPT and PRINCIPLE

3.1 The Limit Definition of the Derivative

The derivative is born from the desire to find the slope of a curve at a single point $x = c$. In algebra, the slope $m$ requires two points: $(x_1, y_1)$ and $(x_2, y_2)$, where $m = \frac{y_2 - y_1}{x_2 - x_1}$.

In Calculus, we take a point $x$ and a slightly shifted point $x+h$. The slope of the secant line through $(x, f(x))$ and $(x+h, f(x+h))$ is: $$m_{sec} = \frac{f(x+h) - f(x)}{(x+h) - x} = \frac{f(x+h) - f(x)}{h}$$

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To find the instantaneous slope, we let the distance $h$ shrink to zero. This leads to the formal definition of the derivative: $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

3.2 Differentiability and Continuity

For a derivative to exist at $x = c$, the limit defined above must exist and be finite. Theorem: If $f$ is differentiable at $x=c$, then $f$ is continuous at $x=c$.

Where Differentiability Fails:

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3.3 Basic Differentiation Rules

To avoid the limit definition for every problem, we derive several fundamental rules:

1. Constant Rule: $\frac{d}{dx}[c] = 0$
2. Constant Multiple Rule: $\frac{d}{dx}[k \cdot f(x)] = k \cdot f'(x)$
3. Sum/Difference Rule: $\frac{d}{dx}[f(x) \pm g(x)] = f'(x) \pm g'(x)$
4. Power Rule: $\frac{d}{dx}[x^n] = n x^{n-1}$ for any real number $n$.

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3.4 Derivatives of Transcendental Functions

The AP curriculum requires memorization and application of the following:

• Exponential: $\frac{d}{dx}[e^x] = e^x$
• Logarithmic: $\frac{d}{dx}[\ln x] = \frac{1}{x}$
• Sine: $\frac{d}{dx}[\sin x] = \cos x$
• Cosine: $\frac{d}{dx}[ \cos x] = -\sin x$

3.5 The Product and Quotient Rules

When functions are multiplied or divided, their derivatives are not simply the product or quotient of their individual derivatives.

The Product Rule: If $h(x) = f(x)g(x)$, then: $$h'(x) = f'(x)g(x) + f(x)g'(x)$$

The Quotient Rule: If $h(x) = \frac{f(x)}{g(x)}$, then: $$h'(x) = \frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2}$$ (Mnemonic: "Low d-High minus High d-Low, over Low-Low")

3.6 Derivatives of Other Trigonometric Functions

Using the Quotient Rule and the identities for $\sin x$ and $\cos x$, we derive:

4. EXAMPLES

Example 1: Limit Definition (Linear)

Find the derivative of $f(x) = 3x + 5$ using the limit definition. $$f'(x) = \lim_{h \to 0} \frac{3(x+h)+5 - (3x+5)}{h} = \lim_{h \to 0} \frac{3x+3h+5-3x-5}{h} = \lim_{h \to 0} \frac{3h}{h} = 3$$

Example 2: Limit Definition (Quadratic)

Find $f'(x)$ for $f(x) = x^2$. $$f'(x) = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h} = \lim_{h \to 0} \frac{x^2+2xh+h^2-x^2}{h} = \lim_{h \to 0} (2x+h) = 2x$$

Example 3: Rational Function via Definition

Find $f'(x)$ for $f(x) = \frac{1}{x}$. $$f'(x) = \lim_{h \to 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h} = \lim_{h \to 0} \frac{\frac{x - (x+h)}{x(x+h)}}{h} = \lim_{h \to 0} \frac{-h}{h x(x+h)} = \lim_{h \to 0} \frac{-1}{x(x+h)} = -\frac{1}{x^2}$$

Example 4: Square Root via Definition

Find $f'(x)$ for $f(x) = \sqrt{x}$. $$f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}$$ Multiply by conjugate: $$\lim_{h \to 0} \frac{(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x})}{h(\sqrt{x+h} + \sqrt{x})} = \lim_{h \to 0} \frac{x+h-x}{h(\sqrt{x+h} + \sqrt{x})} = \frac{1}{2\sqrt{x}}$$

Example 5: Constant Rule

Calculate $\frac{d}{dx} [ \pi^2 ]$. Since $\pi^2$ is a constant, $\frac{d}{dx} [ \pi^2 ] = 0$.

Example 6: Basic Power Rule

Differentiate $f(x) = x^{10}$. $$f'(x) = 10x^{10-1} = 10x^9$$

Example 7: Negative Exponents

Differentiate $f(x) = \frac{1}{x^4}$. Rewrite as $f(x) = x^{-4}$. $$f'(x) = -4x^{-4-1} = -4x^{-5} = -\frac{4}{x^5}$$

Example 8: Fractional Exponents

Differentiate $f(x) = \sqrt[3]{x^2}$. Rewrite as $f(x) = x^{2/3}$. $$f'(x) = \frac{2}{3}x^{2/3 - 1} = \frac{2}{3}x^{-1/3} = \frac{2}{3\sqrt[3]{x}}$$

Example 9: Constant Multiple and Sum Rule

Differentiate $f(x) = 4x^3 - 5x^2 + 7$. $$f'(x) = 4(3x^2) - 5(2x) + 0 = 12x^2 - 10x$$

Example 10: Product Rule

Differentiate $f(x) = x^2 e^x$. Let $u = x^2$ and $v = e^x$. Then $u' = 2x$ and $v' = e^x$. $$f'(x) = (2x)(e^x) + (x^2)(e^x) = e^x(2x + x^2)$$

Example 11: Quotient Rule

Differentiate $f(x) = \frac{\sin x}{x}$. Let $u = \sin x$ and $v = x$. Then $u' = \cos x$ and $v' = 1$. $$f'(x) = \frac{x(\cos x) - (\sin x)(1)}{x^2} = \frac{x\cos x - \sin x}{x^2}$$

Example 12: Equation of a Tangent Line

Find the equation of the tangent line to $f(x) = x^2 + 3x$ at $x = 1$.

1. Point: $f(1) = 1^2 + 3(1) = 4$. Point is $(1, 4)$.
2. Slope: $f'(x) = 2x + 3$. $f'(1) = 2(1) + 3 = 5$.
3. Equation: $y - 4 = 5(x - 1) \implies y = 5x - 1$.

Example 13: Horizontal Tangents

Find the $x$-values where $f(x) = x^3 - 3x^2$ has a horizontal tangent. Horizontal tangents occur where $f'(x) = 0$. $$f'(x) = 3x^2 - 6x = 3x(x - 2)$$ Set to zero: $3x(x - 2) = 0 \implies x = 0, x = 2$.

Example 14: Differentiability Analysis (Piecewise)

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

1. Continuity: $1^2 = 1$ and $2(1)-1 = 1$. Continuous.
2. Left-hand derivative: $\frac{d}{dx}[x^2] = 2x$. At $x=1$, $LHD = 2$.
3. Right-hand derivative: $\frac{d}{dx}[2x-1] = 2$. At $x=1$, $RHD = 2$. Since $LHD = RHD$, $f(x)$ is differentiable at $x = 1$.

Example 15: Differentiability Analysis (Cusp)

Is $f(x) = x^{2/3}$ differentiable at $x = 0$? $f'(x) = \frac{2}{3}x^{-1/3} = \frac{2}{3\sqrt[3]{x}}$. As $x \to 0$, $f'(x) \to \pm \infty$. The derivative is undefined; thus, not differentiable (vertical tangent/cusp).

Example 16: Derivative of $\tan x$

Derive $\frac{d}{dx}[\tan x]$ using $\frac{\sin x}{\cos x}$. $$f'(x) = \frac{\cos x(\cos x) - \sin x(-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x$$

Example 17: Derivative of $\sec x$

Derive $\frac{d}{dx}[\sec x]$ using $\frac{1}{\cos x}$. $$f'(x) = \frac{\cos x(0) - 1(-\sin x)}{\cos^2 x} = \frac{\sin x}{\cos^2 x} = \frac{1}{\cos x} \cdot \frac{\sin x}{\cos x} = \sec x \tan x$$

Example 18: Higher Order Derivatives

Find $f''(x)$ for $f(x) = e^x + x^4$. $f'(x) = e^x + 4x^3$ $f''(x) = e^x + 12x^2$

Example 19: Vertical Tangent Line

Find where $f(x) = \sqrt[3]{x-2}$ has a vertical tangent line. $f'(x) = \frac{1}{3}(x-2)^{-2/3} = \frac{1}{3\sqrt[3]{(x-2)^2}}$. The derivative is undefined when $x-2 = 0 \implies x = 2$. Since $f$ is continuous at $x=2$ and the limit of the derivative is infinite, there is a vertical tangent at $x=2$.

Example 20: Average vs Instantaneous Rate of Change

For $s(t) = t^2$, find the average rate of change on $[1, 3]$ and the instantaneous rate of change at $t = 2$. Average: $\frac{s(3)-s(1)}{3-1} = \frac{9-1}{2} = 4$. Instantaneous: $s'(t) = 2t$. $s'(2) = 2(2) = 4$.

Example 21: Natural Log and Power Rule

Differentiate $y = x^3 \ln x$. $$y' = (3x^2)(\ln x) + (x^3)(\frac{1}{x}) = 3x^2 \ln x + x^2$$

Example 22: Complicated Quotient Rule

Differentiate $f(x) = \frac{e^x}{x^2 + 1}$. $$f'(x) = \frac{(x^2+1)e^x - e^x(2x)}{(x^2+1)^2} = \frac{e^x(x^2 - 2x + 1)}{(x^2+1)^2} = \frac{e^x(x-1)^2}{(x^2+1)^2}$$

Example 23: Derivative of Cosecant

Differentiate $y = \csc x = \frac{1}{\sin x}$. $$y' = \frac{\sin x(0) - 1(\cos x)}{\sin^2 x} = -\frac{\cos x}{\sin^2 x} = -\csc x \cot x$$

Example 24: Finding Constants for Differentiability

Find $a$ and $b$ so $f(x) = \begin{cases} ax^2 + b & x \leq 1 \ \frac{1}{x} & x > 1 \end{cases}$ is differentiable everywhere.

1. Continuity at $x=1$: $a(1)^2 + b = \frac{1}{1} \implies a + b = 1$.
2. Slopes match at $x=1$: $\frac{d}{dx}[ax^2+b] = 2ax$. $\frac{d}{dx}[\frac{1}{x}] = -x^{-2}$. $2a(1) = -(1)^{-2} \implies 2a = -1 \implies a = -1/2$.
3. Substitute $a$: $-1/2 + b = 1 \implies b = 3/2$.

Example 25: Physics Application (Velocity)

A particle moves with position $s(t) = -16t^2 + 64t$. Find its velocity at $t = 1$. $v(t) = s'(t) = -32t + 64$. $v(1) = -32(1) + 64 = 32$ units/sec.

Example 26: Derivative of a Triple Product

Differentiate $f(x) = x e^x \sin x$. Treat as $f(x) = (x e^x) \cdot (\sin x)$. $f'(x) = [\frac{d}{dx}(xe^x)] \sin x + (xe^x) \cos x$ $f'(x) = (1 \cdot e^x + x e^x) \sin x + xe^x \cos x = e^x \sin x + xe^x \sin x + xe^x \cos x$.

Example 27: Estimating Derivative from a Table

Example 28: Limit Definition with Trig

Evaluate $\lim_{h \to 0} \frac{\sin(\frac{\pi}{6} + h) - \sin(\frac{\pi}{6})}{h}$. Recognize this as the definition of $f'(x)$ for $f(x) = \sin x$ at $x = \pi/6$. $f'(x) = \cos x$. $f'(\pi/6) = \cos(\pi/6) = \frac{\sqrt{3}}{2}$.

Example 29: Normal Line Equation

Find the equation of the normal line (perpendicular to tangent) to $y = x^2$ at $x = 3$.

1. Point: $(3, 9)$.
2. Tangent slope: $y' = 2x$, at $x=3$, $m = 6$.
3. Normal slope: $m_{\perp} = -1/6$.
4. Equation: $y - 9 = -\frac{1}{6}(x - 3)$.

Example 30: Second Derivative of $1/x$

Find $\frac{d^2y}{dx^2}$ for $y = x^{-1}$. $\frac{dy}{dx} = -x^{-2}$. $\frac{d^2y}{dx^2} = 2x^{-3} = \frac{2}{x^3}$.

Example 31: Advanced Power Rule derivation (Binomial Theorem approach)

Show $\frac{d}{dx}[x^n] = nx^{n-1}$ for integer $n$ using $\lim_{h \to 0} \frac{(x+h)^n - x^n}{h}$. By Binomial Theorem: $(x+h)^n = x^n + nx^{n-1}h + \frac{n(n-1)}{2}x^{n-2}h^2 + ... + h^n$. $f'(x) = \lim_{h \to 0} \frac{(x^n + nx^{n-1}h + \text{terms with } h^2) - x^n}{h}$ $f'(x) = \lim_{h \to 0} (nx^{n-1} + \text{terms with } h) = nx^{n-1}$.

Example 32: Analyzing a Cusp with Absolute Value

Differentiate $f(x) = |x^2 - 4|$. $f(x) = \begin{cases} x^2 - 4 & |x| \geq 2 \ 4 - x^2 & |x| < 2 \end{cases}$. $f'(x) = \begin{cases} 2x & |x| > 2 \ -2x & |x| < 2 \end{cases}$. At $x = 2$, LHD is $-2(2) = -4$, RHD is $2(2) = 4$. Derivative does not exist at $x = 2, -2$.

Example 33: Tangent line to a Reciprocal function

Find the tangent to $f(x) = \frac{4}{x^2}$ at $x = 2$. $f(x) = 4x^{-2} \implies f'(x) = -8x^{-3} = -\frac{8}{x^3}$. $f'(2) = -\frac{8}{8} = -1$. Point: $f(2) = 1$. Line: $y - 1 = -1(x - 2) \implies y = -x + 3$.

Example 34: Marginal Cost

A cost function is $C(x) = 0.01x^2 + 2x + 100$. Find the marginal cost (derivative) when $x=10$. $C'(x) = 0.02x + 2$. $C'(10) = 0.02(10) + 2 = 2.2$.

Example 35: Limit definition for $e^x$ (Conceptual)

Given $\lim_{h \to 0} \frac{e^h - 1}{h} = 1$, prove $\frac{d}{dx} e^x = e^x$. $$\frac{d}{dx} e^x = \lim_{h \to 0} \frac{e^{x+h} - e^x}{h} = \lim_{h \to 0} \frac{e^x e^h - e^x}{h} = e^x \lim_{h \to 0} \frac{e^h - 1}{h} = e^x(1) = e^x$$