Calculus I Study Guide: Definition of the Derivative and Differentiability

Meaning of the Derivative: Rates of Change and Geometry

• The study of differential calculus begins by analyzing how quantities change over intervals versus how they change at a single precise instant.

• Average Rate of Change (AROC):

  • The average rate of change of a function $f(x)$ over a closed interval $[a, b]$ measures the net change in the function's output values relative to the change in input values.

  • Mathematical definition:     $\text{AROC} = \frac{\Delta y}{\Delta x} = \frac{f(b) - f(a)}{b - a}$

  • Geometric interpretation: The average rate of change is equivalent to the slope $m_{\text{sec}}$ of the secant line passing through two distinct points on the graph of $f(x)$: $(a, f(a))$ and $(b, f(b))$.

• Transitioning from Secant Line to Tangent Line:

  • Let the distance between the two input values $a$ and $b$ be defined as a small nonzero offset $h$. Thus, $b = a + h$.

  • Substituting $b = a + h$ into the secant line slope formula yields the difference quotient:     $m_{\text{sec}} = \frac{f(a + h) - f(a)}{(a + h) - a} = \frac{f(a + h) - f(a)}{h}$

  • Geometrically, as $h$ approaches $0$ ($h \to 0$), the point $(a + h, f(a + h))$ moves continuously along the curve toward the fixed point $(a, f(a))$.

  • As $h \to 0$, the secant line pivots around $(a, f(a))$ and gradually approaches a single limiting line known as the tangent line at $x = a$.

• Instantaneous Rate of Change (IROC) and the Tangent Line Slope:

  • The instantaneous rate of change of $f(x)$ at the specific point $x = a$ is defined as the exact limit of the average rates of change as the interval width $h$ shrinks to zero.

  • Geometric interpretation: The slope of the tangent line $m_{\text{tan}}$ at $x = a$ is the limit of the secant line slopes $m_{\text{sec}}$ as $h \to 0$:     $m_{\text{tan}} = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$

  • The value of this limit is called the derivative of $f(x)$ at $x = a$, denoted as $f'(a)$.

• Conceptual Distinction Between $f'(x)$ and $f'(a)$:

  • $f'(x)$ represents the general derivative function. It accepts an input variable $x$ and outputs a formula that calculates the instantaneous slope at any arbitrary point on $f(x)$.

  • $f'(a)$ represents a single numeric value (a real number) representing the slope of the tangent line at one specific given point $x = a$.

Formal Definition of the Derivative and Difference Quotients

• The General Limit Definition of the Derivative:

  • The derivative of a function $f(x)$ with respect to $x$ is defined by the fundamental limit:     $f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

  • This limit must exist for the derivative to exist at $x$. If the limit does not exist, the function is non-differentiable at $x$.

• The Limit Definition at a Specific Point $x = a$:

  • The derivative at a constant input $x = a$ can be expressed in two mathematically equivalent limit forms:

  1. $h$-formulation:        $f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$

  2. $x$-formulation:        $f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$

• Understanding the Difference Quotient:

  • The expression inside the limit, $\frac{f(x + h) - f(x)}{h}$, is called the difference quotient.

  • Memory Rule for the Difference Quotient: Think of it as "new value minus old value over $h$".

  • New output value: $f(x + h)$ (evaluating the function at a shifted position $x + h$).

  • Old output value: $f(x)$ (the original unmodified function).

  • Change in input: $h$ (the offset distance).

Step-by-Step Limit Definition Procedures by Function Type

• General Algebraic Workflow for Limit Definition Problems:

  • Direct substitution of $h = 0$ into $\frac{f(x + h) - f(x)}{h}$ yields the indeterminate form $\frac{0}{0}$. This indicates that algebraic simplification is required before the limit can be evaluated.

  • Objective: Simplify the numerator until every term without a factor of $h$ cancels out completely, allowing $h$ in the denominator to be factored out and canceled.

  • Standard sequence:

  1. Calculate $f(x + h)$ by replacing every instance of $x$ with $(x + h)$.

  2. Subtract the entire original function $f(x)$, distributing any negative signs across all terms of $f(x)$.

  3. Expand all algebraic expressions and combine like terms in the numerator.

  4. Factor out $h$ from the numerator.

  5. Cancel the factor of $h$ in the numerator with the $h$ in the denominator.

  6. Evaluate the limit by setting $h = 0$ in the simplified expression.

Problem Type 1: Linear Functions

• A. HOW TO RECOGNIZE IT:

  • Problem statement asks to find the derivative using the definition of a linear function of the form $f(x) = m x + b$.

• B. FIRST MOVE:

  • Write out the definition formula: $f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$, then replace $x$ with $(x + h)$ in $f(x)$.

• C. STEPS:

  1. Express $f(x + h) = m (x + h) + b = m x + m h + b$.

  2. Set up the difference quotient: $\frac{(m x + m h + b) - (m x + b)}{h}$.

  3. Distribute the negative sign: $\frac{m x + m h + b - m x - b}{h}$.

  4. Combine like terms in the numerator: $\frac{m h}{h}$.

  5. Cancel $h$: $m$.

  6. Take the limit as $h \to 0$: $\lim_{h \to 0} m = m$.

• D. FULL EXAMPLE:

  • Calculate the derivative of $f(x) = 4x - 7$ using the limit definition.

  • Step 1: Find $f(x + h)$:     $f(x + h) = 4(x + h) - 7 = 4x + 4h - 7$

  • Step 2: Set up the limit definition:     $f'(x) = \lim_{h \to 0} \frac{(4x + 4h - 7) - (4x - 7)}{h}$

  • Step 3: Distribute the negative sign in the numerator:     $f'(x) = \lim_{h \to 0} \frac{4x + 4h - 7 - 4x + 7}{h}$

  • Step 4: Cancel like terms $(4x - 4x = 0$ and $-7 + 7 = 0$):     $f'(x) = \lim_{h \to 0} \frac{4h}{h}$

  • Step 5: Divide out $h$:     $f'(x) = \lim_{h \to 0} 4$

  • Step 6: Evaluate the limit:     $f'(x) = 4$

• E. WHY IT WORKS:

  • A line has a constant slope $m$ everywhere. The tangent line to a straight line is the line itself, so the derivative must equal the constant slope $m$.

• F. COMMON TRAPS:

  • Forgetting to distribute the negative sign to $-7$, writing $-4x - 7$ instead of $-4x + 7$.

• G. MEMORY TRICK:

  • Linear slopes never change; the derivative of $mx + b$ is always just $m$

• H. CHECK YOUR UNDERSTANDING:

  • Problem: Use the definition of the derivative to find $f'(x)$ for $f(x) = -3x + 8$.

  • Solution:     $f'(x) = \lim_{h \to 0} \frac{[-3(x + h) + 8] - [-3x + 8]}{h}$     $f'(x) = \lim_{h \to 0} \frac{-3x - 3h + 8 + 3x - 8}{h}$     $f'(x) = \lim_{h \to 0} \frac{-3h}{h} = \lim_{h \to 0} (-3) = -3$

Problem Type 2: Quadratic Functions (Single Term and Multi-Term)

• A. HOW TO RECOGNIZE IT:

  • Direct instruction to use the limit definition on a quadratic function $f(x) = ax^2 + bx + c$.

• B. FIRST MOVE:

  • Write the formula $f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$.

  • Substitute $(x + h)$ for every $x$, using parentheses carefully.

• C. STEPS:

  1. Expand $(x + h)^2 = x^2 + 2xh + h^2$.

  2. Distribute all coefficients across the expanded polynomial.

  3. Subtract $f(x)$ by distributing the negative sign across all original terms.

  4. Cancel all non-$h$ terms in the numerator.

  5. Factor out $h$ from the remaining numerator terms.

  6. Cancel the $h$ in the numerator with the $h$ in the denominator.

  7. Substitute $h = 0$ into the remaining expression.

• D. FULL EXAMPLE:

  • Calculate the derivative of $f(x) = 3x^2 - 5x + 2$ using the limit definition.

  • Step 1: Write $f(x + h)$:     $f(x + h) = 3(x + h)^2 - 5(x + h) + 2$     $f(x + h) = 3(x^2 + 2xh + h^2) - 5x - 5h + 2$     $f(x + h) = 3x^2 + 6xh + 3h^2 - 5x - 5h + 2$

  • Step 2: Set up difference quotient:     $f'(x) = \lim_{h \to 0} \frac{(3x^2 + 6xh + 3h^2 - 5x - 5h + 2) - (3x^2 - 5x + 2)}{h}$

  • Step 3: Distribute the negative sign:     $f'(x) = \lim_{h \to 0} \frac{3x^2 + 6xh + 3h^2 - 5x - 5h + 2 - 3x^2 + 5x - 2}{h}$

  • Step 4: Cancel terms without $h$ $(3x^2 - 3x^2 = 0$, $-5x + 5x = 0$, $2 - 2 = 0$):     $f'(x) = \lim_{h \to 0} \frac{6xh + 3h^2 - 5h}{h}$

  • Step 5: Factor out $h$ from numerator:     $f'(x) = \lim_{h \to 0} \frac{h(6x + 3h - 5)}{h}$

  • Step 6: Cancel $h$:     $f'(x) = \lim_{h \to 0} (6x + 3h - 5)$

  • Step 7: Substitute $h = 0$:     $f'(x) = 6x + 3(0) - 5 = 6x - 5$

• E. WHY IT WORKS:

  • Expanding $(x+h)^2$ creates terms containing $h$ and $h^2$. Subtracting $f(x)$ eliminates all static terms without $h$. Factoring out $h$ removes the $\frac{0}{0}$ indeterminate singularity, making direct substitution of $h=0$ possible.

• F. COMMON TRAPS:

  • Writing $(x + h)^2 = x^2 + h^2$ (forgetting the middle term $2xh$).

  • Failing to distribute the negative sign across the entire subtractive term $-(3x^2 - 5x + 2)$.

• G. MEMORY TRICK:

  • Expand, distribute, cancel non-$h$s, factor $h$, reduce, and sub $h=0$

• H. CHECK YOUR UNDERSTANDING:

  • Problem: Use the definition of the derivative to calculate $f'(x)$ for $f(x) = -2x^2 + 4x$.

  • Solution:     $f(x + h) = -2(x + h)^2 + 4(x + h) = -2(x^2 + 2xh + h^2) + 4x + 4h = -2x^2 - 4xh - 2h^2 + 4x + 4h$     $f'(x) = \lim_{h \to 0} \frac{-2x^2 - 4xh - 2h^2 + 4x + 4h - (-2x^2 + 4x)}{h}$     $f'(x) = \lim_{h \to 0} \frac{-2x^2 - 4xh - 2h^2 + 4x + 4h + 2x^2 - 4x}{h}$     $f'(x) = \lim_{h \to 0} \frac{-4xh - 2h^2 + 4h}{h} = \lim_{h \to 0} \frac{h(-4x - 2h + 4)}{h}$     $f'(x) = \lim_{h \to 0} (-4x - 2h + 4) = -4x + 4$

Problem Type 3: Cubic Functions

• A. HOW TO RECOGNIZE IT:

  • Asked to compute derivative using the limit definition for $f(x) = x^3$ (or multi-term cubics).

• B. FIRST MOVE:

  • State the Binomial Theorem expansion for $(x + h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$.

• C. STEPS:

  1. Expand $f(x + h)$.

  2. Set up Difference Quotient.

  3. Subtract $f(x)$ and simplify numerator.

  4. Factor out $h$ from the remaining terms ($3x^2h + 3xh^2 + h^3$).

  5. Cancel $h$ with the denominator.

  6. Substitute $h = 0$ into the remaining terms.

• D. FULL EXAMPLE:

  • Compute $f'(x)$ for $f(x) = x^3$ using the limit definition.

  • Step 1: Expand $f(x + h)$:     $f(x + h) = (x + h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$

  • Step 2: Set up difference quotient:     $f'(x) = \lim_{h \to 0} \frac{(x^3 + 3x^2h + 3xh^2 + h^3) - x^3}{h}$

  • Step 3: Cancel $x^3 - x^3$:     $f'(x) = \lim_{h \to 0} \frac{3x^2h + 3xh^2 + h^3}{h}$

  • Step 4: Factor out $h$:     $f'(x) = \lim_{h \to 0} \frac{h(3x^2 + 3xh + h^2)}{h}$

  • Step 5: Cancel $h$:     $f'(x) = \lim_{h \to 0} (3x^2 + 3xh + h^2)$

  • Step 6: Set $h = 0$:     $f'(x) = 3x^2 + 3x(0) + (0)^2 = 3x^2$

• E. WHY IT WORKS:

  • Pascal's Triangle row 3 coefficients $(1, 3, 3, 1)$ define the expansion of $(x+h)^3$. The single leading term $x^3$ cancels with $-f(x)$, leaving all terms with powers of $h$, enabling cancellation of the denominator.

• F. COMMON TRAPS:

  • Misexpanding $(x + h)^3$ as $x^3 + h^3$ or misplacing the coefficients $3x^2h$ and $3xh^2$

• G. MEMORY TRICK:

  • $(x+h)^3$ expands to $1, 3, 3, 1$; drop $x^3$ and divide out $h$

• H. CHECK YOUR UNDERSTANDING:

  • Problem: Find $f'(x)$ using the definition of derivative for $f(x) = x^3 + 2x$

  • Solution:     $f(x + h) = (x + h)^3 + 2(x + h) = x^3 + 3x^2h + 3xh^2 + h^3 + 2x + 2h$     $f'(x) = \lim_{h \to 0} \frac{(x^3 + 3x^2h + 3xh^2 + h^3 + 2x + 2h) - (x^3 + 2x)}{h}$     $f'(x) = \lim_{h \to 0} \frac{3x^2h + 3xh^2 + h^3 + 2h}{h} = \lim_{h \to 0} \frac{h(3x^2 + 3xh + h^2 + 2)}{h}$     $f'(x) = \lim_{h \to 0} (3x^2 + 3xh + h^2 + 2) = 3x^2 + 2$

Problem Type 4: Reciprocal / Rational Functions

• A. HOW TO RECOGNIZE IT:

  • Asked to evaluate derivative via definition for rational expressions such as $f(x) = \frac{1}{x}$ or $f(x) = \frac{1}{x + c}$.

• B. FIRST MOVE:

  • Write difference quotient with rational fractions in the numerator:     $\lim_{h \to 0} \frac{\frac{1}{x + h} - \frac{1}{x}}{h}$

• C. STEPS:

  1. Combine fractions in the numerator using a common denominator $x(x + h)$.

  2. Simplify the combined numerator.

  3. Convert the complex fraction into a single fraction by multiplying by $\frac{1}{h}$.

  4. Cancel $h$ in numerator and denominator.

  5. Take limit by substituting $h = 0$

• D. FULL EXAMPLE:

  • Find $f'(x)$ using the limit definition for $f(x) = \frac{1}{x}$.

  • Step 1: Write initial limit expression:     $f'(x) = \lim_{h \to 0} \frac{\frac{1}{x + h} - \frac{1}{x}}{h}$

  • Step 2: Combine numerator over common denominator $x(x + h)$:     $\frac{1}{x + h} - \frac{1}{x} = \frac{x - (x + h)}{x(x + h)} = \frac{x - x - h}{x(x + h)} = \frac{-h}{x(x + h)}$

  • Step 3: Rewrite difference quotient as a single fraction:     $f'(x) = \lim_{h \to 0} \frac{\frac{-h}{x(x + h)}}{h} = \lim_{h \to 0} \left( \frac{-h}{x(x + h)} \cdot \frac{1}{h} \right)$

  • Step 4: Cancel $h$:     $f'(x) = \lim_{h \to 0} \frac{-1}{x(x + h)}$

  • Step 5: Substitute $h = 0$:     $f'(x) = \frac{-1}{x(x + 0)} = -\frac{1}{x^2}$

• E. WHY IT WORKS:

  • Complex fractions cannot be evaluated directly when $h = 0$. Finding a common denominator consolidates all instances of $h$ into a single factor in the main numerator, enabling exact cancellation with the denominator's $h$.

• F. COMMON TRAPS:

  • Subtracting incorrectly in numerator: writing $x - x + h = h$ instead of $x - (x + h) = -h$.

  • Dropping the main denominator $h$ during fraction arithmetic steps.

• G. MEMORY TRICK:

  • Cross-multiply the inner numerators, consolidate denominators, cancel $h$

• H. CHECK YOUR UNDERSTANDING:

  • Problem: Find the derivative of $f(x) = \frac{1}{x + 3}$ using the definition.

  • Solution:     $f'(x) = \lim_{h \to 0} \frac{\frac{1}{x + h + 3} - \frac{1}{x + 3}}{h}$     $f'(x) = \lim_{h \to 0} \frac{\frac{(x + 3) - (x + h + 3)}{(x + h + 3)(x + 3)}}{h} = \lim_{h \to 0} \frac{\frac{-h}{(x + h + 3)(x + 3)}}{h}$     $f'(x) = \lim_{h \to 0} \frac{-h}{h(x + h + 3)(x + 3)} = \lim_{h \to 0} \frac{-1}{(x + h + 3)(x + 3)} = -\frac{1}{(x + 3)^2}$

Problem Type 5: Radical / Square Root Functions

• A. HOW TO RECOGNIZE IT:

  • Asked to compute derivative via definition for root functions like $f(x) = \sqrt{x}$.

• B. FIRST MOVE:

  • Write difference quotient $\lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h}$, then identify the conjugate of the numerator: $(\sqrt{x + h} + \sqrt{x})$.

• C. STEPS:

  1. Multiply both numerator and denominator by the conjugate $(\sqrt{x + h} + \sqrt{x})$.

  2. Use the difference of squares identity $(A - B)(A + B) = A^2 - B^2$ to expand the numerator.

  3. Simplify the numerator: (\sqrt{x + h})^2 - (\sqrt{x})^2 = (x + h) - x = h$.\n 4. Do **not** expand the denominator; keep it as h(\sqrt{x + h} + \sqrt{x}).\n 5. Cancel h$in the numerator with$h in the denominator.\n 6. Evaluate the limit by setting h = 0.\n\n- **D. FULL EXAMPLE:**\n - Calculate derivative of f(x) = \sqrt{x} using the definition.\n - Step 1: Write limit setup:\n    f'(x) = \lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h}\n - Step 2: Multiply numerator and denominator by conjugate:\n    f'(x) = \lim_{h \to 0} \frac{(\sqrt{x + h} - \sqrt{x})(\sqrt{x + h} + \sqrt{x})}{h(\sqrt{x + h} + \sqrt{x})}\n - Step 3: Expand numerator using difference of squares:\n    f'(x) = \lim_{h \to 0} \frac{(x + h) - x}{h(\sqrt{x + h} + \sqrt{x})} = \lim_{h \to 0} \frac{h}{h(\sqrt{x + h} + \sqrt{x})}\n - Step 4: Cancel h:\n    f'(x) = \lim_{h \to 0} \frac{1}{\sqrt{x + h} + \sqrt{x}}\n - Step 5: Substitute h = 0:\n    f'(x) = \frac{1}{\sqrt{x + 0} + \sqrt{x}} = \frac{1}{\sqrt{x} + \sqrt{x}} = \frac{1}{2\sqrt{x}}\n\n- **E. WHY IT WORKS:**\n - Radical expressions cannot be added or subtracted directly under roots. Conjugate multiplication eliminates radical symbols from the numerator using difference of squares, exposing h directly for factor cancellation.\n\n- **F. COMMON TRAPS:**\n - Distributing h$into the denominator conjugate expression. Keep denominator factored as$h(\text{conjugate}).\n - Concluding that \sqrt{x} + \sqrt{x} = \sqrt{2x}$instead of$2\sqrt{x}.\n\n- **G. MEMORY TRICK:**\n - Conjugate multiplication turns square roots into simple terms\n\n- **H. CHECK YOUR UNDERSTANDING:**\n - *Problem:* Find f'(x)$using the definition for$f(x) = \sqrt{x + 5}.\n - *Solution:* \n    f'(x) = \lim_{h \to 0} \frac{\sqrt{x + h + 5} - \sqrt{x + 5}}{h}\n    f'(x) = \lim_{h \to 0} \frac{(\sqrt{x + h + 5} - \sqrt{x + 5})(\sqrt{x + h + 5} + \sqrt{x + 5})}{h(\sqrt{x + h + 5} + \sqrt{x + 5})}\n    f'(x) = \lim_{h \to 0} \frac{(x + h + 5) - (x + 5)}{h(\sqrt{x + h + 5} + \sqrt{x + 5})} = \lim_{h \to 0} \frac{h}{h(\sqrt{x + h + 5} + \sqrt{x + 5})}\n    f'(x) = \lim_{h \to 0} \frac{1}{\sqrt{x + h + 5} + \sqrt{x + 5}} = \frac{1}{2\sqrt{x + 5}}\n\n\n# Derivative at a Specific Point\n\n- **Two Methodologies for Computing f'(a):**\n - **Method 1 (General First, Evaluate Second):** Compute the general derivative expression f'(x)$using limits or derivative rules, then substitute$x = a.\n - **Method 2 (Direct Point Limit):** Plug x = a directly into the definition limit from the start:\n    f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}\n\n- **Comparative Walkthrough Example:** Find f'(3)$for$f(x) = x^2 - 4x\n - *Using Method 1:* \n 1. Find general f'(x):\n       f'(x) = \lim_{h \to 0} \frac{[(x+h)^2 - 4(x+h)] - [x^2 - 4x]}{h} = 2x - 4\n 2. Plug in a = 3:\n       f'(3) = 2(3) - 4 = 2\n - *Using Method 2:* \n 1. Calculate f(3) = (3)^2 - 4(3) = 9 - 12 = -3\n 2. Calculate f(3 + h) = (3 + h)^2 - 4(3 + h) = 9 + 6h + h^2 - 12 - 4h = h^2 + 2h - 3\n 3. Set up point limit:\n       f'(3) = \lim_{h \to 0} \frac{(h^2 + 2h - 3) - (-3)}{h} = \lim_{h \to 0} \frac{h^2 + 2h}{h} = \lim_{h \to 0} (h + 2) = 2\n\n- **Key Distinction:** Method 1 produces a general continuous slope function f'(x) = 2x - 4$valid for all$x$. Method 2 computes only the single static real number$2$, representing the slope specifically at$x = 3.\n\n\n# Tangent-Line Equations and Geometry\n\n- **A. HOW TO RECOGNIZE IT:**\n - Problem asks to "find the equation of the line tangent to f(x)$at$x = a" or "find the tangent line equation."\n\n- **B. FIRST MOVE:**\n - State the Four-Step Tangent Line Procedure: **Derivative \to$Slope$\to$Point$\to Point-Slope Equation**.\n\n- **C. STEPS:**\n 1. **Derivative:** Find f'(x).\n 2. **Slope (m$):** Substitute$x = a$into$f'(x)$to calculate$m = f'(a).\n 3. **Point (y_1$):** Substitute$x = a$into the **original** function$f(x)$to calculate$y_1 = f(a)$. The point is$(a, f(a)).\n 4. **Equation:** Plug m$,$x_1 = a$, and$y_1 = f(a) into point-slope form:\n     y - y_1 = m(x - x_1)\n     Optionally rewrite in slope-intercept form (y = mx + b).\n\n- **D. FULL EXAMPLE:**\n - Find the equation of the tangent line to f(x) = 2x^2 - 3x + 1$at$x = 2\n - Step 1: Find derivative f'(x):\n    f'(x) = 4x - 3\n - Step 2: Calculate slope m$at$a = 2:\n    m = f'(2) = 4(2) - 3 = 8 - 3 = 5\n - Step 3: Calculate y_1$coordinate using original$f(x):\n    y_1 = f(2) = 2(2)^2 - 3(2) + 1 = 2(4) - 6 + 1 = 3\n    The point of tangency is (2, 3).\n - Step 4: Assemble point-slope equation:\n    y - 3 = 5(x - 2)\n    y - 3 = 5x - 10\n    y = 5x - 7\n\n- **E. WHY IT WORKS:**\n - A tangent line touches the curve at (a, f(a))$and matches its instantaneous rate of change at that point. Thus, its linear slope must equal$f'(a)$and it must pass through the point$(a, f(a)) on the original graph.\n\n- **F. COMMON TRAPS:**\n - Plugging x = a$into the derivative$f'(x)$to find the$y_1$coordinate instead of using the original function$f(x).\n - Plugging x = a$into the original function$f(x)$to find the slope$m$instead of using$f'(x).\n\n- **G. MEMORY TRICK:**\n - Derivative gives slope m$; Original gives point$y\n\n- **H. CHECK YOUR UNDERSTANDING:**\n - *Problem:* Find the equation of the tangent line to f(x) = x^3 - 2x$at$x = -1\n - *Solution:* \n 1. Derivative: f'(x) = 3x^2 - 2\n 2. Slope: m = f'(-1) = 3(-1)^2 - 2 = 3(1) - 2 = 1\n 3. Point: y_1 = f(-1) = (-1)^3 - 2(-1) = -1 + 2 = 1$. Point is$(-1, 1).\n 4. Equation: y - 1 = 1(x - (-1)) \implies y - 1 = x + 1 \implies y = x + 2\n\n\n# The Derivative as a Function and Notation Systems\n\n- **Conceptual Shift from Section 3.1 to Section 3.2:**\n - In Section 3.1, the derivative is introduced as a local tool to find the slope at one specified point x = a.\n - In Section 3.2, we generalize this concept: by letting the point vary across all values of x$in the domain, the derivative itself becomes a standalone **function**$f'(x).\n\n- **Complete Derivative Notations:**\n - **Prime Notation (Lagrange):**\n - f'(x)$(read as "$f$prime of$x")\n - y'$(read as "$y prime")\n - **Leibniz Notation:**\n - \frac{dy}{dx}$(read as "the derivative of$y$with respect to$x")\n - \frac{df}{dx}$(read as "the derivative of$f$with respect to$x")\n - \frac{d}{dx}[f(x)]$(read as "$d$$d$$x$of$f(x)")\n - **Operator Notation (Euler / Cauchy):**\n - D f(x)$or$D_x f(x)$(read as "the differential operator with respect to$x$applied to$f(x)")\n - **Evaluation Notation (Leibniz at a point):**\n - \left.\frac{dy}{dx}\right|_{x=a}$(read as "$d$$y$over$d$$x$evaluated at$x$equals$a")\n\n\n# The Power Rule for Differentiation\n\n- **The Power Rule Formula:**\n - For any real number constant exponent n:\n    \frac{d}{dx}\left(x^n\right) = n x^{n-1}\n\n- **Memory Rule:**\n - **"Bring the exponent down in front as a coefficient, and subtract one from the exponent."**\n\n- **Application Across Exponent Types:**\n - **Positive Integers:** \n    \frac{d}{dx}\left(x^5\right) = 5x^{5-1} = 5x^4\n - **Negative Exponents:** \n    \frac{d}{dx}\left(x^{-3}\right) = -3 x^{-3-1} = -3 x^{-4} = -\frac{3}{x^4}\n - **Fractional / Rational Exponents:** \n    \frac{d}{dx}\left(x^{1/2}\right) = \frac{1}{2} x^{(1/2)-1} = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}}\n - **Unit Exponent (n = 1):** \n    \frac{d}{dx}(x) = 1 \cdot x^{1-1} = 1 \cdot x^0 = 1 \cdot 1 = 1\n - **Zero Exponent / Constants (n = 0):** \n    \frac{d}{dx}(c) = \frac{d}{dx}\left(c \cdot x^0\right) = c \cdot 0 \cdot x^{-1} = 0\n\n\n# Linearity Rules of Differentiation\n\n- **1. Constant Rule:**\n - The derivative of any constant number is always zero:\n    \frac{d}{dx}(c) = 0\n - *Geometric reason:* The graph of y = c$is a flat horizontal line with slope$0 everywhere.\n\n- **2. Power Rule for x:**\n - The derivative of x is always one:\n    \frac{d}{dx}(x) = 1\n - *Geometric reason:* The graph of y = x$is a straight line with constant slope$1\n\n- **3. Constant Multiple Rule:**\n - A constant factor multiplicative coefficient carries through unchanged during differentiation:\n    \frac{d}{dx}[c \cdot f(x)] = c \cdot f'(x)\n - *Example:* \frac{d}{dx}\left(7x^4\right) = 7 \cdot \frac{d}{dx}\left(x^4\right) = 7 \cdot (4x^3) = 28x^3\n\n- **4. Sum Rule:**\n - The derivative of a sum is the sum of individual derivatives:\n    \frac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x)\n\n- **5. Difference Rule:**\n - The derivative of a difference is the difference of individual derivatives:\n    \frac{d}{dx}[f(x) - g(x)] = f'(x) - g'(x)\n\n- **Term-by-Term Polynomial Differentiation Example:**\n - Differentiate P(x) = 4x^5 - 3x^3 + 8x^2 - 6x + 12\n    P'(x) = 4(5x^4) - 3(3x^2) + 8(2x) - 6(1) + 0\n    P'(x) = 20x^4 - 9x^2 + 16x - 6\n\n\n# Algebraic Pre-Processing: Rewrite BEFORE Differentiating\n\n- **Crucial Rule:** Power Rule directly applies **only** to base x$raised to a constant power ($x^n). Expressions with radicals or variables in denominators MUST be rewritten algebraically before applying derivative rules.\n\n- **Algebraic Rewrite Conversion Identities:**\n - **Reciprocal Conversion:** \n    \frac{1}{x^n} = x^{-n}\n - **Radical Conversion:** \n    \sqrt[n]{x^m} = x^{m/n}\n - **Combined Reciprocal-Radical Conversion:** \n    \frac{1}{\sqrt[n]{x^m}} = x^{-m/n}\n - **Numerator Division Conversion:** \n    \frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}\n\n- **Three-Step Workflow: REWRITE \to$DIFFERENTIATE$\to SIMPLIFY**\n\n- **Comprehensive Pre-Processing Examples:**\n\n - **Example A: Reciprocal Term**\n - Function: f(x) = \frac{5}{x^3}\n - 1. Rewrite: f(x) = 5 x^{-3}\n - 2. Differentiate: f'(x) = 5 (-3 x^{-4}) = -15 x^{-4}\n - 3. Simplify: f'(x) = -\frac{15}{x^4}\n\n - **Example B: Radical Term**\n - Function: f(x) = 4\sqrt[3]{x^2}\n - 1. Rewrite: f(x) = 4 x^{2/3}\n - 2. Differentiate: f'(x) = 4 \left(\frac{2}{3} x^{(2/3)-1}\right) = \frac{8}{3} x^{-1/3}\n - 3. Simplify: f'(x) = \frac{8}{3 x^{1/3}} = \frac{8}{3\sqrt[3]{x}}\n\n - **Example C: Fraction Splitting**\n - Function: f(x) = \frac{x^4 - 3\sqrt{x}}{x}\n - 1. Rewrite (split terms and apply fraction rules):\n       f(x) = \frac{x^4}{x} - \frac{3 x^{1/2}}{x^1} = x^3 - 3 x^{(1/2)-1} = x^3 - 3 x^{-1/2}\n - 2. Differentiate:\n       f'(x) = 3x^2 - 3 \left(-\frac{1}{2} x^{-3/2}\right) = 3x^2 + \frac{3}{2} x^{-3/2}\n - 3. Simplify:\n       f'(x) = 3x^2 + \frac{3}{2 x^{3/2}} = 3x^2 + \frac{3}{2\sqrt{x^3}}\n\n\n# The Natural Exponential Function Rule\n\n- **Derivative of the Natural Exponential Function:**\n - The function e^x is its own derivative:\n    \frac{d}{dx}\left(e^x\right) = e^x\n\n- **Constant Multiple with e^x:**\n  \frac{d}{dx}\left(c e^x\right) = c e^x\n\n- **Linear Combinations Involving Exponential Functions:**\n - Differentiate f(x) = 3e^x - 2x^4 + \frac{1}{x}\n - Step 1 (Rewrite): f(x) = 3e^x - 2x^4 + x^{-1}\n - Step 2 (Differentiate): f'(x) = 3e^x - 2(4x^3) + (-1 x^{-2})\n - Step 3 (Simplify): f'(x) = 3e^x - 8x^3 - \frac{1}{x^2}\n\n\n# Evaluating Derivatives at Specific Values\n\n- **Order of Operations Requirement:**\n - **ALWAYS differentiate FIRST to obtain f'(x), then substitute the numeric value SECOND.**\n - *Warning:* Substituting the numeric value before differentiating yields a constant value, whose derivative is incorrectly evaluated as 0\n\n- **Sign Rules when Substituting Negative Numbers:**\n - When evaluating expressions like x^2$or$x^4$with a negative input$x = -a, keep parentheses around the base:\n    (-3)^2 = 9 \quad \text{whereas} \quad -3^2 = -9\n - Negative exponents: \n    (-2)^{-3} = \frac{1}{(-2)^3} = \frac{1}{-8} = -\frac{1}{8}\n\n- **Full Walkthrough Example:**\n - Evaluate f'(-2)$for$f(x) = x^4 - 3x^2 + 5x\n 1. Differentiate first:\n       f'(x) = 4x^3 - 6x + 5\n 2. Substitute x = -2 using explicit parentheses:\n       f'(-2) = 4(-2)^3 - 6(-2) + 5\n       f'(-2) = 4(-8) + 12 + 5 = -32 + 12 + 5 = -15\n\n\n# Finding Horizontal Tangent Lines\n\n- **A. HOW TO RECOGNIZE IT:**\n - Question asks to "find all values of x" or "find all points where the graph has a horizontal tangent line."\n\n- **B. FIRST MOVE:**\n - A horizontal line has slope m = 0$. Set the derivative equal to zero: **$f'(x) = 0**.\n\n- **C. STEPS:**\n 1. Find f'(x).\n 2. Set f'(x) = 0\n 3. Solve the algebraic equation completely for all real roots x\n 4. Read the problem carefully:\n - If it asks for **x--values**, stop here.\n - If it asks for **points**, plug each x$--value back into the **original function$f(x)$** to calculate the corresponding$y$--coordinate:$(x, f(x)).\n\n- **D. FULL EXAMPLE:**\n - Find all points on the graph of f(x) = x^3 - 3x^2 - 9x + 5 where the tangent line is horizontal.\n - Step 1: Differentiate:\n    f'(x) = 3x^2 - 6x - 9\n - Step 2: Set f'(x) = 0:\n    3x^2 - 6x - 9 = 0\n - Step 3: Solve for x by factoring:\n    3(x^2 - 2x - 3) = 0\n    3(x - 3)(x + 1) = 0\n    x = 3 \quad \text{or} \quad x = -1\n - Step 4: Find corresponding y$--coordinates using original$f(x):\n - For x = 3:\n      f(3) = (3)^3 - 3(3)^2 - 9(3) + 5 = 27 - 27 - 27 + 5 = -22\n      First point: (3, -22)\n - For x = -1:\n      f(-1) = (-1)^3 - 3(-1)^2 - 9(-1) + 5 = -1 - 3(1) + 9 + 5 = 10\n      Second point: (-1, 10)\n\n- **E. WHY IT WORKS:**\n - A horizontal line has zero steepness (m = 0$). Since the derivative gives the slope of the tangent line, horizontal tangents occur precisely where$f'(x) = 0\n\n- **F. COMMON TRAPS:**\n - Plugging x$--values into$f'(x)$to find$y$--coordinates (which gives$0 instead of the graph position).\n - Stopping at x$--values when the problem asks for coordinate points$(x, y).\n - Forgetting negative roots when factoring or taking square roots.\n\n- **G. MEMORY TRICK:**\n - Horizontal means slope zero: set derivative to 0, solve x$, plug into original for$y\n\n- **H. CHECK YOUR UNDERSTANDING:**\n - *Problem:* Find all points on f(x) = 2x^3 - 6x where the tangent line is horizontal.\n - *Solution:* \n 1. f'(x) = 6x^2 - 6\n 2. 6x^2 - 6 = 0 \implies 6(x^2 - 1) = 0 \implies 6(x - 1)(x + 1) = 0 \implies x = 1, x = -1\n 3. Calculate y--coordinates:\n - f(1) = 2(1)^3 - 6(1) = 2 - 6 = -4 \implies (1, -4)\n - f(-1) = 2(-1)^3 - 6(-1) = -2 + 6 = 4 \implies (-1, 4)\n - Points: (1, -4)$and$(-1, 4)\n\n\n# Continuity and Differentiability Relationships\n\n- **Differentiability Implies Continuity Theorem:**\n - If a function f(x)$is differentiable at$x = a$, then$f(x)$MUST also be continuous at$x = a\n - Formal implication: \n    \text{Differentiable at } x = a \implies \text{Continuous at } x = a\n\n- **Logical Direction and Misconceptions (The Converse Fallacy):**\n - The converse statement is **FALSE**: Continuity does **NOT** imply differentiability.\n - A function can be continuous at a point x = a while failing to be differentiable at that point.\n\n- **Classic Counterexample: The Absolute Value Function f(x) = |x|$at$x = 0:**\n - *Continuity check at x = 0$:*     $\lim_{x \to 0} |x| = 0 = f(0)$     $f(x) = |x|$ is continuous at $x = 0$

  • *Differentiability check at x = 0$:*\n - Left-hand derivative slope:\n      \lim_{h \to 0^-} \frac{|0 + h| - |0|}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1\n - Right-hand derivative slope:\n      \lim_{h \to 0^+} \frac{|0 + h| - |0|}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1\n - Since the left-hand limit (-1$) does not equal the right-hand limit ($1$), the overall limit does not exist. Thus,$f(x) = |x|$is **not differentiable** at$x = 0\n\n- **Four Primary Causes of Non-Differentiability at x = a:**\n 1. **Corner / Sharp Turn:** Left-hand limit and right-hand limit of the difference quotient exist but are unequal (e.g., f(x) = |x|$at$x = 0).\n 2. **Cusp:** Slope approaches \infty$from one side and$-\infty$from the other side (e.g.,$f(x) = x^{2/3}$at$x = 0).\n 3. **Vertical Tangent Line:** Derivative limit approaches \pm \infty$from both sides (e.g.,$f(x) = x^{1/3}$at$x = 0).\n 4. **Discontinuity:** Any jump, removable, or infinite discontinuity automatically makes a function non-differentiable.\n\n\n# Integrated Problem-Solving: Connecting Definitions, Rules, and Tangents\n\n- **Unified Problem Workflow:**\n - Many calculus problems combine multiple concepts into a sequential multi-step problem.\n\n- **Integrated Walkthrough Example:**\n - Consider the function f(x) = x^2 - 3x$.

  • Part A: Use the limit definition of the derivative to find $f'(x)$.     $f'(x) = \lim_{h \to 0} \frac{[(x+h)^2 - 3(x+h)] - [x^2 - 3x]}{h}$     $f'(x) = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - 3x - 3h - x^2 + 3x}{h}$     $f'(x) = \lim_{h \to 0} \frac{2xh + h^2 - 3h}{h} = \lim_{h \to 0} (2x + h - 3) = 2x - 3$

  • Part B: Verify $f'(x)$ using shortcut derivative rules.     $\frac{d}{dx}(x^2 - 3x) = 2x^{2-1} - 3(1) = 2x - 3$

  • Part C: Calculate the instantaneous rate of change at x = 4$.\n    f'(4) = 2(4) - 3 = 8 - 3 = 5\n - **Part D:** Find the equation of the tangent line at x = 4$.

  • Slope: $m = f'(4) = 5$

  • Point: $y_1 = f(4) = (4)^2 - 3(4) = 16 - 12 = 4$. Point is $(4, 4)$.

  • Tangent line: $y - 4 = 5(x - 4) \implies y - 4 = 5x - 20 \implies y = 5x - 16$

Mastery Challenge Problems

Challenge Problem 1: Multi-Term Limit Definition

• Problem: Compute the derivative of $f(x) = 2x^2 - 7x + 4$ using the limit definition.

• Solution Step-by-Step:

  1. Set up limit:      $f'(x) = \lim_{h \to 0} \frac{[2(x + h)^2 - 7(x + h) + 4] - [2x^2 - 7x + 4]}{h}$

  2. Expand terms:      $f'(x) = \lim_{h \to 0} \frac{[2(x^2 + 2xh + h^2) - 7x - 7h + 4] - [2x^2 - 7x + 4]}{h}$      $f'(x) = \lim_{h \to 0} \frac{2x^2 + 4xh + 2h^2 - 7x - 7h + 4 - 2x^2 + 7x - 4}{h}$

  3. Cancel non-$h$ terms:      $f'(x) = \lim_{h \to 0} \frac{4xh + 2h^2 - 7h}{h}$

  4. Factor out $h$ and reduce:      $f'(x) = \lim_{h \to 0} \frac{h(4x + 2h - 7)}{h} = \lim_{h \to 0} (4x + 2h - 7)$

  5. Evaluate limit at $h = 0$:      $f'(x) = 4x - 7$

Challenge Problem 2: Pre-Algebraic Rationalization and Power Rule

• Problem: Find $\frac{dy}{dx}$ for $y = \frac{x^3 + 4x\sqrt{x} - 2}{x}$.

• Solution Step-by-Step:

  1. Rewrite terms individually by dividing each numerator term by denominator $x^1$:      $y = \frac{x^3}{x^1} + \frac{4x^1 x^{1/2}}{x^1} - \frac{2}{x^1}$      $y = x^2 + 4x^{1/2} - 2x^{-1}$

  2. Differentiate term-by-term using Power Rule:      $\frac{dy}{dx} = 2x^{2-1} + 4 \left(\frac{1}{2} x^{(1/2)-1}\right) - 2(-1 x^{-1-1})$      $\frac{dy}{dx} = 2x + 2x^{-1/2} + 2x^{-2}$

  3. Simplify into radical/positive fraction format:      $\frac{dy}{dx} = 2x + \frac{2}{\sqrt{x}} + \frac{2}{x^2}$

Challenge Problem 3: Multi-Point Horizontal Tangent Analysis

• Problem: Find all coordinate points on $f(x) = 2x^3 - 3x^2 - 12x + 7$ where the tangent line is horizontal.

• Solution Step-by-Step:

  1. Calculate derivative $f'(x)$:      $f'(x) = 6x^2 - 6x - 12$

  2. Set $f'(x) = 0$:      $6x^2 - 6x - 12 = 0$

  3. Factor quadratic equation:      $6(x^2 - x - 2) = 0$      $6(x - 2)(x + 1) = 0 \implies x = 2 \quad \text{and} \quad x = -1$

  4. Substitute $x = 2$ into original $f(x)$:      $f(2) = 2(2)^3 - 3(2)^2 - 12(2) + 7 = 16 - 12 - 24 + 7 = -13$      First point: $(2, -13)$

  5. Substitute $x = -1$ into original $f(x)$:      $f(-1) = 2(-1)^3 - 3(-1)^2 - 12(-1) + 7 = -2 - 3 + 12 + 7 = 14$      Second point: $(-1, 14)$