Lecture 12: Instantaneous Rate

The derivative — capturing "how fast, right now?"

The Problem: You're running a reaction. You measure concentration every 10 seconds. Your advisor asks: "What's the reaction rate at t = 25 seconds?"

You don't have a measurement at t = 25. And even if you did — the rate is changing. It's not the same at t = 25 as at t = 30.

How do you answer?

This is not a calculus exercise. This is your actual problem in the lab.

1. The Real Question

Chemistry is full of "how fast?" questions:

Question What you're really asking
"What's the reaction rate?" How fast is concentration changing?
"What's the current?" How fast is charge flowing?
"What's the velocity?" How fast is position changing?
"What's the heat flow?" How fast is energy transferring?

The pattern: You have a quantity that depends on time (or position, or something else). You want to know how fast it's changing at a specific instant.

This is the derivative.

2. The Problem with Average Rate

2.1 Average Rate of Change

Definition: The average rate of change of f over [a, b] is:

$$\text{Average rate} = \frac{f(b) - f(a)}{b - a} = \frac{\Delta f}{\Delta x}$$

This is the slope of the secant line connecting (a, f(a)) and (b, f(b)).

Interactive: Secant Line → Tangent Line

1.0
1.00
Secant Slope
—
True Derivative
—
Error
—

As h → 0, the secant line approaches the tangent line, and the average rate approaches the instantaneous rate.

3. The Derivative: Definition

3.1 Limit Definition

Definition: The derivative of f at x = a is:

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

if this limit exists.

3.2 Notation

Notation Read as Origin
f'(x) "f prime of x" Lagrange
df/dx "dee f dee x" Leibniz
d/dx f(x) "dee dee x of f" Operator form
ḟ "f dot" Newton (time)

Leibniz notation df/dx is powerful because it reminds you: the derivative is a ratio of infinitesimal changes.

4. Computing Derivatives from the Definition

Example 1: f(x) = x²

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

Example 2: f(x) = eˣ

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

The exponential function is its own derivative! This is why eˣ is special.

Interactive: The Derivative as Slope

0.0
f(x) = x²
0.00
f'(x) = 2x
0.00
Behavior
Minimum

5. What the Derivative Tells You

5.1 Geometric Meaning

5.2 Physical Meaning

If f(t) represents... Then f'(t) is...
Position Velocity
Velocity Acceleration
Concentration Reaction rate
Charge Current
Energy Power

6. Chemistry Connection: Reaction Rates

For a reaction A → Products:

$$\text{Rate} = -\frac{d[A]}{dt}$$

Interactive: First-Order Kinetics

0.10
10
[A] at time t
—
Rate = -d[A]/dt
—
Verification: k[A]
—

First-order kinetics: Rate = k[A]. The rate is the derivative!

7. Chemistry Connection: Force from Potential

If V(x) is the potential energy as a function of position:

$$F = -\frac{dV}{dx}$$

Interactive: Force from Potential Energy

0.5
V(x)
—
F = -dV/dx
—
Direction
—

8. When the Derivative Doesn't Exist

The derivative f'(a) exists only if the limit exists (finite, same from both sides).

Interactive: Points Where Derivative Fails

Left derivative
-1
Right derivative
+1
Status
Not equal!

Key theorem: Differentiable ⟹ Continuous (but not vice versa)

9. Basic Derivatives Table

f(x) f'(x) Notes
c (constant) 0 Constants don't change
xⁿ nxⁿ⁻¹ Power rule
eˣ eˣ Its own derivative!
ln x 1/x
sin x cos x
cos x -sin x

10. Exercises

Exercise 1: From the Definition

Use the limit definition to find f'(x) for:

  1. f(x) = 3x + 2
  2. f(x) = x³
  3. f(x) = √x
  1. f'(x) = lim[(3(x+h)+2) - (3x+2)]/h = lim 3h/h = 3
  2. f'(x) = lim[(x+h)³ - x³]/h = lim[3x²h + 3xh² + h³]/h = 3x²
  3. f'(x) = lim[√(x+h) - √x]/h · (conjugate) = lim 1/(√(x+h) + √x) = 1/(2√x)

Exercise 2: Reaction Rate

A reaction has concentration profile [A](t) = 0.5e^(-0.1t) M.

  1. What is the rate at t = 0?
  2. What is the rate at t = 10 s?
  3. At what time is the rate equal to 0.01 M/s?

d[A]/dt = -0.05e^(-0.1t), so Rate = |d[A]/dt| = 0.05e^(-0.1t)

  1. At t = 0: Rate = 0.05e^0 = 0.05 M/s
  2. At t = 10: Rate = 0.05e^(-1) = 0.0184 M/s
  3. 0.01 = 0.05e^(-0.1t) → t = 16.1 s

Exercise 3: Force from Potential

A particle moves in the potential V(x) = x⁴ - 2x².

  1. Find F(x) = -dV/dx
  2. Where is F = 0? (Equilibrium points)
  3. Which equilibrium points are stable?
  1. F(x) = -(4x³ - 4x) = 4x(1 - x²) = 4x - 4x³
  2. F = 0: 4x(1-x²) = 0 → x = 0, ±1
  3. d²V/dx² = 12x² - 4. At x=0: -4 < 0 (unstable). At x=±1: 8 > 0 (stable)

Exercise 4: Linear Approximation

  1. Use linear approximation to estimate e^(0.1)
  2. Use linear approximation to estimate sin(0.1)
  1. f(x) = eˣ, f'(x) = eˣ, f(0) = 1, f'(0) = 1 → e^0.1 ≈ 1 + 0.1 = 1.1 (actual: 1.1052)
  2. f(x) = sin x, f'(x) = cos x, f(0) = 0, f'(0) = 1 → sin(0.1) ≈ 0 + 0.1 = 0.1 (actual: 0.0998)

Coming Up: Lecture 13

Rules of Change — The power rule, product rule, chain rule

Computing derivatives efficiently without limits every time

References: McQuarrie, Mathematics for Physical Chemistry; Atkins, Physical Chemistry