Chapter 5 • Theory & Derivations

The Derivative: Foundations, Difference Quotients & Differentiation Rules

Rigorous theory of differentiation: secant-to-tangent limiting process, Newton difference quotient, one-sided derivatives, proof that differentiability implies continuity, failure modes (cusps, vertical tangents, wild oscillations), power rule, product rule proof, quotient rule, and derivatives of trigonometric, exponential, and hyperbolic functions.

§5.1 The Tangent Line Problem, Instantaneous Rates & Difference Quotients

1. The Tangent Line as a Limiting Secant Line

Let $P(x_0, f(x_0))$ be a fixed point on the curve $y = f(x)$. Let $Q(x_0 + h, f(x_0 + h))$ be a neighboring point with $h \ne 0$. The slope of the secant line passing through $P$ and $Q$ is given by the Newton-Leibniz difference quotient:

m_{\text{sec}} = \frac{\Delta y}{\Delta x} = \frac{f(x_0 + h) - f(x_0)}{h}

As $h \to 0$, point $Q$ slides along the curve toward $P$. If the secant slopes approach a unique finite limiting value, that limit is defined as the slope of the tangent line to the curve at $P$:

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

2. Equations of Tangent and Normal Lines

Tangent Line: Passing through $(x_0, y_0)$ with slope $m = f'(x_0)$: $$$y - f(x_0) = f'(x_0)(x - x_0)$$$
Normal Line: The line perpendicular to the tangent line at $(x_0, y_0)$ (provided $f'(x_0) \ne 0$, having slope $m_{\perp} = -1/f'(x_0)$): $y - f(x_0) = -\frac{1}{f'(x_0)}(x - x_0)$

§5.2 One-Sided Derivatives & Differentiability Implying Continuity

1. One-Sided Derivatives

\begin{aligned} \mathbf{\text{Right-Hand Derivative: }} & f'_+(x_0) = \lim_{h \to 0^+} \frac{f(x_0 + h) - f(x_0)}{h} \ \mathbf{\text{Left-Hand Derivative: }} & f'_-(x_0) = \lim_{h \to 0^-} \frac{f(x_0 + h) - f(x_0)}{h} \end{aligned}

A function is differentiable at $x_0$ if and only if both one-sided derivatives exist as finite numbers and are equal: $f'(x_0) = f'_+(x_0) = f'_-(x_0)$.

2. Theorem: Differentiability Implies Continuity

\mathbf{\text{Theorem: If } f \text{ is differentiable at } x_0, \text{ then } f \text{ is continuous at } x_0.}

Proof: To prove continuity, we must show that $\lim_{x \to x_0} [f(x) - f(x_0)] = 0$. For $x \ne x_0$, write:

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

Taking the limit as $x \to x_0$ using the product limit law:

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

Thus $\lim_{x \to x_0} f(x) = f(x_0)$, proving that $f$ is continuous at $x_0 \quad \blacksquare$

3. The Converse is False: Geometric Non-Differentiability Modes

Continuity is a strictly weaker condition than differentiability. Functions may fail to be differentiable at points where they are continuous due to:

Corner Point (Cusp): Left and right derivatives differ ($f(x) = |x|$ at $x = 0$, where $f'_-(0) = -1 \ne f'_+(0) = +1$).
Vertical Tangent: The difference quotient diverges to $\pm\infty$ ($f(x) = x^{1/3}$ at $x = 0$, where $f'(0) = \lim_{h \to 0} h^{1/3}/h = \lim_{h \to 0} 1/h^{2/3} = \infty$).
Wild Oscillation: $f(x) = x \sin(1/x)$ for $x \ne 0$ and $f(0) = 0$ is continuous at $x = 0$, but its difference quotient $\sin(1/h)$ oscillates indefinitely between $-1$ and $+1$ as $h \to 0$.

§5.3 Fundamental Algebraic Differentiation Rules & Rigorous Proofs

1. The Power Rule

\mathbf{\frac{d}{dx}[x^n] = n x^{n-1} \quad \forall n \in \mathbb{R}}

Proof for $n \in \mathbb{N}$ via Binomial Theorem:

\begin{aligned} \frac{d}{dx}[x^n] &= \lim_{h \to 0} \frac{(x + h)^n - x^n}{h} = \lim_{h \to 0} \frac{\left( x^n + n x^{n-1} h + \binom{n}{2} x^{n-2} h^2 + \dots + h^n \right) - x^n}{h} \ &= \lim_{h \to 0} \left( n x^{n-1} + \binom{n}{2} x^{n-2} h + \dots + h^{n-1} \right) = n x^{n-1} \quad \blacksquare \end{aligned}

2. Linearity of the Derivative Operator

\frac{d}{dx}[c f(x)] = c f'(x), \qquad \frac{d}{dx}[f(x) \pm g(x)] = f'(x) \pm g'(x)

3. The Product Rule & Line-by-Line Deductive Proof

\mathbf{\frac{d}{dx}[f(x) g(x)] = f'(x) g(x) + f(x) g'(x)}

Proof: Form the difference quotient for the product $P(x) = f(x) g(x)$:

\frac{P(x + h) - P(x)}{h} = \frac{f(x + h) g(x + h) - f(x) g(x)}{h}

Add and subtract the strategic intermediate term $f(x) g(x + h)$ in the numerator:

\begin{aligned} &= \frac{f(x + h) g(x + h) - f(x) g(x + h) + f(x) g(x + h) - f(x) g(x)}{h} \ &= \left( \frac{f(x + h) - f(x)}{h} \right) g(x + h) + f(x) \left( \frac{g(x + h) - g(x)}{h} \right) \end{aligned}

Taking the limit as $h \to 0$: since $g$ is differentiable at $x$, it is continuous at $x$, so $\lim_{h \to 0} g(x + h) = g(x)$. Therefore:

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

4. The Quotient Rule

\mathbf{\frac{d}{dx}\left[ \frac{f(x)}{g(x)} \right] = \frac{f'(x) g(x) - f(x) g'(x)}{[g(x)]^2} \quad (g(x) \ne 0)}

§5.4 Derivatives of Trigonometric, Exponential & Hyperbolic Functions

1. Derivatives of Trigonometric Functions

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

\begin{aligned} \frac{d}{dx}[\sin x] &= \lim_{h \to 0} \frac{\sin(x + h) - \sin x}{h} = \lim_{h \to 0} \frac{\sin x(\cos h - 1) + \cos x\sin h}{h} \ &= \sin x \left( \lim_{h \to 0} \frac{\cos h - 1}{h} \right) + \cos x \left( \lim_{h \to 0} \frac{\sin h}{h} \right) = \sin x(0) + \cos x(1) = \cos x \end{aligned}

Similarly, using the Quotient Rule:

\frac{d}{dx}[\cos x] = -\sin x, \quad \frac{d}{dx}[\tan x] = \sec^2 x, \quad \frac{d}{dx}[\sec x] = \sec x \tan x, \quad \frac{d}{dx}[\cot x] = -\csc^2 x, \quad \frac{d}{dx}[\csc x] = -\csc x \cot x

2. Derivatives of Exponential & Logarithmic Functions

\frac{d}{dx}[e^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 (1) = e^x$$ $$\frac{d}{dx}[a^x] = a^x \ln a, \qquad \frac{d}{dx}[\ln x] = \frac{1}{x} \quad (x > 0), \qquad \frac{d}{dx}[\log_a x] = \frac{1}{x \ln a}

3. Derivatives of Hyperbolic Functions

\frac{d}{dx}[\sinh x] = \frac{d}{dx}\left[\frac{e^x - e^{-x}}{2}\right] = \frac{e^x + e^{-x}}{2} = \cosh x$$ $$\frac{d}{dx}[\cosh x] = \sinh x, \qquad \frac{d}{dx}[\tanh x] = \operatorname{sech}^2 x, \qquad \frac{d}{dx}[\operatorname{sech} x] = -\operatorname{sech} x \tanh x

TIERED UNIVERSITY HONORS PROBLEMS

Step-by-Step Solved Examination Problems

Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.

Tier 1: Foundational Example 5.1: Derivatives via Product and Quotient Rules

Compute the exact derivative $f'(x)$ for: (a) $f(x) = (3x^2 - 5x + 2) e^x$, and (b) $g(x) = \frac{x^2 \sin x}{x + \cos x}$.

Step 1: Apply Product Rule to (a)

$$f'(x) = \frac{d}{dx}[3x^2 - 5x + 2] e^x + (3x^2 - 5x + 2) \frac{d}{dx}[e^x] = (6x - 5) e^x + (3x^2 - 5x + 2) e^x$$

Differentiating term by term and factoring out $e^x$.

Step 2: Simplify (a)

$$f'(x) = (3x^2 + x - 3) e^x$$

Combining like algebraic polynomial coefficients.

Step 3: Apply Quotient Rule to (b)

$$g'(x) = \frac{\frac{d}{dx}[x^2 \sin x] (x + \cos x) - (x^2 \sin x) \frac{d}{dx}[x + \cos x]}{(x + \cos x)^2}$$

Compute numerator derivative: $\frac{d}{dx}[x^2 \sin x] = 2x \sin x + x^2 \cos x$. Compute denominator derivative: $\frac{d}{dx}[x + \cos x] = 1 - \sin x$.

Step 4: Expand and Assemble (b)

$$g'(x) = \frac{(2x \sin x + x^2 \cos x)(x + \cos x) - x^2 \sin x (1 - \sin x)}{(x + \cos x)^2}$$

This gives the exact derivative.

Final Answer & Physical Insight

f'(x) = (3x^2 + x - 3) e^x, \qquad g'(x) = \frac{(2x \sin x + x^2 \cos x)(x + \cos x) - x^2 \sin x (1 - \sin x)}{(x + \cos x)^2}

Tier 2: Intermediate Exam Example 5.2: Tangent and Normal Line Equations to a Transcendental Curve

Find the exact Cartesian equations of the tangent line and the normal line to the curve $y = f(x) = x^2 \ln(x) - 3x$ at the point where $x = 1$.

Step 1: Compute Point Coordinates $(x_0, y_0)$

$$y_0 = f(1) = 1^2 \ln(1) - 3(1) = 0 - 3 = -3$$

The point of tangency on the curve is $(1, -3)$.

Step 2: Compute Derivative $f'(x)$

$$f'(x) = \frac{d}{dx}[x^2 \ln x] - \frac{d}{dx}[3x] = \left( 2x \ln x + x^2 \cdot \frac{1}{x} \right) - 3 = 2x \ln x + x - 3$$

Applying the Product Rule to $x^2 \ln x$.

Step 3: Evaluate Tangent Slope at $x = 1$

$$m_{\text{tan}} = f'(1) = 2(1) \ln(1) + 1 - 3 = 0 + 1 - 3 = -2$$

The tangent slope is $m = -2$. The normal slope is $m_{\perp} = -1/(-2) = 1/2$.

Step 4: Formulate Line Equations

$$\text{Tangent: } y - (-3) = -2(x - 1) \implies y = -2x - 1 \\ \text{Normal: } y - (-3) = \frac{1}{2}(x - 1) \implies y = \frac{1}{2}x - \frac{7}{2}$$

Writing in slope-intercept form.

Final Answer & Physical Insight

\text{Tangent Line: } y = -2x - 1, \qquad \text{Normal Line: } y = \frac{1}{2}x - \frac{7}{2}

Tier 3: Honors / Proof Challenge Example 5.3: Differentiability vs Continuity Analysis of an Oscillatory Piecewise Function

Consider the function $f(x) = \begin{cases} x^2 \sin\left(\frac{1}{x}\right), & x \ne 0 \\ 0, & x = 0 \end{cases}$. (a) Prove using the limit definition that $f'(0)$ exists and find its value. (b) For $x \ne 0$, compute $f'(x)$. (c) Prove that $\lim_{x \to 0} f'(x)$ does NOT exist, and explain why this demonstrates that a derivative function need not be continuous.

Step 1: Evaluate $f'(0)$ from the Definition of Derivative

$$f'(0) = \lim_{h \to 0} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0} \frac{h^2 \sin(1/h) - 0}{h} = \lim_{h \to 0} h \sin\left(\frac{1}{h}\right)$$

We must evaluate this limit as $h o 0$.

Step 2: Apply the Squeeze Theorem to $f'(0)$

$$-1 \le \sin(1/h) \le 1 \implies -|h| \le h \sin(1/h) \le |h| \implies \lim_{h \to 0} h \sin(1/h) = 0$$

Because $\lim_{h o 0} (\pm|h|) = 0$, the Squeeze Theorem guarantees that $f'(0) = 0$ exists.

Step 3: Compute $f'(x)$ for $x e 0$

$$f'(x) = \frac{d}{dx}[x^2] \sin(1/x) + x^2 \frac{d}{dx}[\sin(1/x)] = 2x \sin\left(\frac{1}{x}\right) + x^2 \cos\left(\frac{1}{x}\right)\left(-\frac{1}{x^2}\right) = 2x \sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right)$$

Applying product and chain rules for $x e 0$.

Step 4: Prove Non-Existence of $\lim_{x o 0} f'(x)$

$$\lim_{x \to 0} f'(x) = \lim_{x \to 0} \left[ 2x \sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right) \right] = 0 - \lim_{x \to 0} \cos\left(\frac{1}{x}\right) \quad (\text{DNE})$$

As $x o 0$, $2x \sin(1/x) o 0$ by squeezing, but $\cos(1/x)$ oscillates between $-1$ and $+1$ infinitely often. Thus $\lim_{x o 0} f'(x)$ does not exist. Since $\lim_{x o 0} f'(x) e f'(0) = 0$, the derivative $f'(x)$ is discontinuous at $x = 0$, providing a classic counterexample.

Final Answer & Physical Insight

f'(0) = 0 \text{ exists}; \quad f'(x) = 2x\sin(1/x) - \cos(1/x) \text{ for } x \ne 0; \quad \lim_{x \to 0} f'(x) \text{ DNE (discontinuous derivative)}