Chapter 2: Differential Calculus – Master Derivatives

Derivatives are the cornerstone of differential calculus, describing how a function changes at a specific point. They answer questions like: How fast is a car accelerating? or What's the slope of a curve at a given point? This chapter builds from the definition of derivatives to their practical applications, with step-by-step examples to solidify your understanding.

2.1 Definition of Derivatives

The derivative captures the "instantaneous rate of change" of a function. To understand it, we start with the average rate of change and refine it to the instantaneous case using limits.

2.1.1 The Limit-Based Definition

For a function $f(x)$, the derivative at a point $x = a$, denoted $f'(a)$ (read "f prime of a"), is defined as:

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

This is called the difference quotient limit. Here, $h$ represents a small change in $x$ (from $a$ to $a + h$), and the numerator is the corresponding change in $f(x)$. As $h$ approaches 0, the average rate of change over the interval $[a, a + h]$ becomes the instantaneous rate of change at $x = a$.

2.1.2 Geometric Meaning: Slope of the Tangent Line

The derivative $f'(a)$ equals the slope of the tangent line to the graph of $f(x)$ at the point $(a, f(a))$.

• The tangent line is the line that "just touches" the curve at $(a, f(a))$ and has the same steepness as the curve at that point.
• Contrast this with the secant line, which connects two points on the curve $(a, f(a))$ and $(a + h, f(a + h))$; its slope is the average rate of change $\frac{f(a + h) - f(a)}{h}$. As $h \to 0$, the secant line approaches the tangent line.

Example 1: Compute $f'(2)$ for $f(x) = x^2$ using the definition

Step 1: Substitute into the difference quotient:

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

Step 2: Evaluate $f(2 + h)$ and $f(2)$:

$$f(2 + h) = (2 + h)^2 = 4 + 4h + h^2 \\ f(2) = 2^2 = 4$$

Step 3: Simplify the numerator:

$$f(2 + h) - f(2) = (4 + 4h + h^2) - 4 = 4h + h^2$$

Step 4: Divide by $h$ and take the limit:

$$f'(2) = \lim_{h \to 0} \frac{4h + h^2}{h} = \lim_{h \to 0} (4 + h) = 4$$

Thus, the slope of the tangent line to $f(x) = x^2$ at $x = 2$ is $4$.

---

2.1.3 Physical Meaning: Instantaneous Velocity

In physics, if $s(t)$ represents the position of an object at time $t$, then:

• The average velocity over $[t, t + h]$ is $\frac{s(t + h) - s(t)}{h}$
• The instantaneous velocity at time $t$ is the derivative $s'(t)$, the limit of the average velocity as $h \to 0$

Example 2: Instantaneous velocity of a falling object

A ball is dropped from a height, and its position (in meters) at time $t$ (in seconds) is $s(t) = -4.9t^2 + 50$ (negative because it falls downward). Find its velocity at $t = 2$ seconds.

Step 1: Use the derivative definition for $s'(2)$:

$$s'(2) = \lim_{h \to 0} \frac{s(2 + h) - s(2)}{h}$$

Step 2: Compute $s(2 + h)$ and $s(2)$:

$$\begin{align*} s(2 + h) &= -4.9(2 + h)^2 + 50 \\ &= -4.9(4 + 4h + h^2) + 50 \\ &= -19.6 - 19.6h - 4.9h^2 + 50 \\ &= 30.4 - 19.6h - 4.9h^2 \\ s(2) &= -4.9(2)^2 + 50 = -19.6 + 50 = 30.4 \end{align*}$$

Step 3: Simplify and take the limit:

$$\begin{align*} s'(2) &= \lim_{h \to 0} \frac{(30.4 - 19.6h - 4.9h^2) - 30.4}{h} \\ &= \lim_{h \to 0} \frac{-19.6h - 4.9h^2}{h} \\ &= \lim_{h \to 0} (-19.6 - 4.9h) = -19.6 \end{align*}$$

The velocity at $t = 2$ is $-19.6 \text{ m/s}$ (negative indicates downward motion).

2.2 Derivative Calculation Rules

Calculating derivatives using the limit definition is tedious. Instead, we use rules derived from the definition to simplify computations.

2.2.1 Basic Rules

Let $f(x)$ and $g(x)$ be differentiable functions, and $c$ be a constant.

Rule	Formula	Example
Constant Rule	$\frac{d}{dx}[c] = 0$	$\frac{d}{dx}[5] = 0$
Power Rule	$\frac{d}{dx}[x^n] = nx^{n-1}$	$\frac{d}{dx}[x^3] = 3x^2$; $\frac{d}{dx}[x^{1/2}] = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$
Constant Multiple Rule	$\frac{d}{dx}[cf(x)] = c \cdot f'(x)$	$\frac{d}{dx}[3x^2] = 3 \cdot 2x = 6x$
Sum Rule	$\frac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x)$	$\frac{d}{dx}[x^2 + \sin x] = 2x + \cos x$
Difference Rule	$\frac{d}{dx}[f(x) - g(x)] = f'(x) - g'(x)$	$\frac{d}{dx}[e^x - \ln x] = e^x - \frac{1}{x}$

2.2.2 Product Rule

For the product of two functions:

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

Example: Find $\frac{d}{dx}[x^2 \cdot \sin x]$

Let $f(x) = x^2$ (so $f'(x) = 2x$) and $g(x) = \sin x$ (so $g'(x) = \cos x$).

Apply the product rule:

$$\frac{d}{dx}[x^2 \cdot \sin x] = 2x \cdot \sin x + x^2 \cdot \cos x$$

2.2.3 Quotient Rule

For the quotient of two functions ($g(x) \neq 0$):

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

Example: Find $\frac{d}{dx}\left[\frac{x}{x + 1}\right]$

Let $f(x) = x$ (so $f'(x) = 1$) and $g(x) = x + 1$ (so $g'(x) = 1$).

Apply the quotient rule:

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

2.2.4 Chain Rule (Critical for Composite Functions)

Composite functions are "functions within functions," e.g., $h(x) = \sin(x^2)$ (inner function $x^2$, outer function $\sin(\cdot)$). The chain rule states:

If $h(x) = f(g(x))$, then:

$$h'(x) = f'(g(x)) \cdot g'(x)$$

In words: "Derivative of the outer function, evaluated at the inner function, multiplied by the derivative of the inner function."

Example 1: Simple composite function

Find $\frac{d}{dx}[\sin(x^2)]$

• Outer function: $f(u) = \sin(u)$, so $f'(u) = \cos(u)$
• Inner function: $u = g(x) = x^2$, so $g'(x) = 2x$

Apply the chain rule:

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

Example 2: Nested composite function

Find $\frac{d}{dx}[\sqrt{1 + e^{3x}}]$

Rewrite as $(1 + e^{3x})^{1/2}$:

• Outer function: $f(u) = u^{1/2}$, so $f'(u) = \frac{1}{2}u^{-1/2}$
• Middle function: $u = g(v) = 1 + e^v$, so $g'(v) = e^v$
• Inner function: $v = h(x) = 3x$, so $h'(x) = 3$

Apply the chain rule (twice, for nested functions):

$$\frac{d}{dx}[\sqrt{1 + e^{3x}}] = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)$$

Substitute:

$$= \frac{1}{2}(1 + e^{3x})^{-1/2} \cdot e^{3x} \cdot 3 = \frac{3e^{3x}}{2\sqrt{1 + e^{3x}}}$$

2.2.5 Derivatives of Common Functions

Memorize these to speed up calculations (derivations use limits or rules above):

Function $f(x)$	Derivative $f'(x)$
$e^x$	$e^x$ (unique: derivative equals itself)
$a^x$ ($a > 0, a \neq 1$)	$a^x \ln a$
$\ln x$ ($x > 0$)	$\frac{1}{x}$
$\log_a x$ ($x > 0$)	$\frac{1}{x \ln a}$
$\sin x$	$\cos x$
$\cos x$	$-\sin x$
$\tan x$	$\sec^2 x$

2.3 Applications of Derivatives

Derivatives solve real-world problems involving change, optimization, and more.

2.3.1 Function Monotonicity (Increasing/Decreasing Intervals)

A function $f(x)$ is:

• Increasing on an interval if $f'(x) > 0$ for all $x$ in the interval
• Decreasing on an interval if $f'(x) < 0$ for all $x$ in the interval

Example: Find where $f(x) = x^3 - 3x$ is increasing/decreasing

Step 1: Find the derivative: $f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x - 1)(x + 1)$

Step 2: Find critical points (where $f'(x) = 0$ or undefined): $3(x - 1)(x + 1) = 0 \rightarrow x = 1 \text{ or } x = -1$

Step 3: Test intervals divided by critical points:

• For $x < -1$ (e.g., $x = -2$): $f'(-2) = 3(3)(-1) = 9 > 0$ → increasing
• For $-1 < x < 1$ (e.g., $x = 0$): $f'(0) = 3(-1)(1) = -3 < 0$ → decreasing
• For $x > 1$ (e.g., $x = 2$): $f'(2) = 3(1)(3) = 9 > 0$ → increasing

Conclusion: $f(x)$ increases on $(-\infty, -1) \cup (1, +\infty)$ and decreases on $(-1, 1)$

2.3.2 Extremum and Maximum/Minimum Values

• Critical points: $x = c$ where $f'(c) = 0$ or $f'(c)$ is undefined (candidates for extrema)
• Relative (local) extremum: A peak or valley in the graph. Use the First Derivative Test to classify:
  • If $f'(x)$ changes from positive to negative at $x = c$, $f(c)$ is a local maximum
  • If $f'(x)$ changes from negative to positive at $x = c$, $f(c)$ is a local minimum

Example: Find local extrema of $f(x) = x^3 - 3x$ (from the previous example)

Critical points: $x = -1$ and $x = 1$

• At $x = -1$: $f'(x)$ changes from positive (left of -1) to negative (right of -1) → local maximum $f(-1) = (-1)^3 - 3(-1) = -1 + 3 = 2$

• At $x = 1$: $f'(x)$ changes from negative (left of 1) to positive (right of 1) → local minimum $f(1) = (1)^3 - 3(1) = 1 - 3 = -2$

2.3.3 Practical Optimization

Derivatives help maximize/minimize quantities like profit, area, or material usage.

Example: Minimize material for a box

A box with an open top has a volume of $108 \text{ cm}^3$. Find the dimensions that minimize the surface area.

Step 1: Define variables. Let the base length/width be $x$ (square base for symmetry) and height be $h$

Step 2: Volume constraint: $\text{Volume} = x \cdot x \cdot h = x^2h = 108 \rightarrow h = \frac{108}{x^2}$

Step 3: Surface area (open top: base + 4 sides): $A = x^2 + 4(xh)$ Substitute $h$: $A(x) = x^2 + 4\left(x \cdot \frac{108}{x^2}\right) = x^2 + \frac{432}{x}$

Step 4: Find critical points. Compute $A'(x)$: $A'(x) = 2x - \frac{432}{x^2}$ Set to 0: $2x = \frac{432}{x^2} \rightarrow 2x^3 = 432 \rightarrow x^3 = 216 \rightarrow x = 6 \text{ cm}$

Step 5: Find $h$: $h = \frac{108}{6^2} = \frac{108}{36} = 3 \text{ cm}$

Conclusion: Dimensions $6 \text{ cm} \times 6 \text{ cm} \times 3 \text{ cm}$ minimize surface area

2.3.4 Related Rates

These problems involve variables changing with time, where their rates of change are related.

Example: Water filling a tank

A cylindrical tank with radius 5 m is filled with water at a rate of $3 \text{ m}^3/\text{min}$. How fast is the water level rising?

Step 1: Variables. Let $h(t)$ = height of water at time $t$; volume $V(t) = \pi r^2 h$ (cylinder volume). Here, $r = 5 \text{ m}$ (constant), so $V = 25\pi h$

Step 2: Differentiate both sides with respect to $t$ (chain rule for related rates): $\frac{dV}{dt} = 25\pi \cdot \frac{dh}{dt}$

Step 3: Substitute known rate: $\frac{dV}{dt} = 3 \text{ m}^3/\text{min}$. Solve for $\frac{dh}{dt}$: $3 = 25\pi \cdot \frac{dh}{dt} \rightarrow \frac{dh}{dt} = \frac{3}{25\pi} \approx 0.038 \text{ m/min}$

2.4 Practice Problems (with Answer Hints)

1. Use the limit definition to find $f'(1)$ for $f(x) = 2x + 3$
   Hint: The difference quotient simplifies to a constant

2. Compute $\frac{d}{dx}[x^4 \cos x]$ using the product rule

3. Find the derivative of $h(x) = \ln(x^2 + 1)$ using the chain rule

4. Determine where $f(x) = x^2 - 6x + 5$ is increasing/decreasing

5. A ladder 10 ft long leans against a wall. If the bottom slides away from the wall at 2 ft/s, how fast is the top sliding down when the bottom is 6 ft from the wall?

Answer Hints

1. $2$ (the function is linear, so the derivative is its slope)
2. $4x^3 \cos x - x^4 \sin x$
3. $\frac{2x}{x^2 + 1}$
4. Decreasing on $(-\infty, 3)$, increasing on $(3, +\infty)$ (critical point at $x = 3$)
5. $-1.5 \text{ ft/s}$ (negative indicates downward motion)