Lecture 13: Rules of Change

Differentiation rules — computing derivatives efficiently

The Problem: The rate of an enzyme-catalyzed reaction:

$$v = \frac{V_{max}[S]}{K_m + [S]}$$

You need dv/d[S] — how sensitive is the rate to substrate concentration?

Using the limit definition would take a page of algebra. There's a better way.

1. The Strategy

Every complicated function is built from simple functions via:

If we know derivatives of simple functions + rules for combining them → we can differentiate anything.

Function Structure Rule to Use
f(x) + g(x) Sum rule
c · f(x) Constant multiple
f(x) · g(x) Product rule
f(x) / g(x) Quotient rule
f(g(x)) Chain rule

2. Basic Derivatives

These you must know cold:

f(x) f'(x) Why
c 0 Constants don't change
xⁿ nxⁿ⁻¹ Power rule
eˣ eˣ Definition of e
ln x 1/x Inverse of eˣ
sin x cos x From definition
cos x -sin x From definition

3. The Power Rule

Power Rule

$$\frac{d}{dx}[x^n] = nx^{n-1}$$

Valid for any real n (positive, negative, fractional).

Interactive: Power Rule

2.0
Function
x²
Derivative
2x

4. Linearity Rules

Constant Multiple Rule

$$\frac{d}{dx}[cf(x)] = c\frac{df}{dx}$$

Sum Rule

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

Combined: Differentiation is a linear operator:

$$\frac{d}{dx}[af(x) + bg(x)] = af'(x) + bg'(x)$$

Example

$$\frac{d}{dx}[3x^4 - 5x^2 + 2x - 7] = 12x^3 - 10x + 2$$

5. The Product Rule

Product Rule

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

Mnemonic: "First times derivative of second, plus second times derivative of first."

Interactive: Product Rule — x²eˣ

1.0
f = x²
1.00
g = eˣ
2.72
f'g + fg'
8.15

6. The Quotient Rule

Quotient Rule

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

Mnemonic: "Low d-high minus high d-low, over low squared."

Example: Michaelis-Menten Sensitivity

$$v = \frac{V_{max}[S]}{K_m + [S]}$$

$$\frac{dv}{d[S]} = \frac{V_{max}(K_m + [S]) - V_{max}[S] \cdot 1}{(K_m + [S])^2} = \frac{V_{max}K_m}{(K_m + [S])^2}$$

7. The Chain Rule

Chain Rule — The Big One

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

In Leibniz notation: If y = f(u) and u = g(x):

$$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$

Intuition: Rates multiply through the chain. If u changes at rate du/dx and y changes at rate dy/du, then y changes at rate (dy/du)(du/dx).

Interactive: Chain Rule — e^(x²)

1.0
Inner: g(x) = x²
1.00
Inner deriv: 2x
2.00
Chain: f'(g)·g'
5.44

More Chain Rule Examples

f(x) Outer Inner f'(x)
(x² + 1)¹⁰ u¹⁰ x² + 1 10(x² + 1)⁹ · 2x
sin(3x + 1) sin u 3x + 1 cos(3x + 1) · 3
ln(x² + 1) ln u x² + 1 2x/(x² + 1)
√(1 - x²) u^(1/2) 1 - x² -x/√(1 - x²)

8. Chemistry: Arrhenius Equation

Rate constant temperature dependence:

$$k = Ae^{-E_a/RT}$$

Using the chain rule:

$$\frac{dk}{dT} = A \cdot e^{-E_a/RT} \cdot \frac{E_a}{RT^2} = \frac{kE_a}{RT^2}$$

Interactive: Arrhenius Temperature Dependence

50
300
k(T)
—
dk/dT
—
% change per K
—

9. Chemistry: Michaelis-Menten Sensitivity

Interactive: Rate Sensitivity to Substrate

20
20
v
—
dv/d[S]
—
At [S]=Km
—

Key insight: As [S] → ∞, dv/d[S] → 0. The enzyme becomes saturated.

10. Summary: All Rules

Rule Formula
Constant (c)' = 0
Power (xⁿ)' = nxⁿ⁻¹
Constant multiple (cf)' = cf'
Sum (f + g)' = f' + g'
Product (fg)' = f'g + fg'
Quotient (f/g)' = (f'g - fg')/g²
Chain [f(g)]' = f'(g) · g'

11. Exercises

Exercise 1: Basic Rules

Differentiate:

  1. f(x) = 5x⁴ - 3x² + 2x - 7
  2. f(x) = √x + 1/x²
  3. f(x) = (x² - 1)(x³ + 2)
  1. f'(x) = 20x³ - 6x + 2
  2. f(x) = x^(1/2) + x^(-2), f'(x) = 1/(2√x) - 2/x³
  3. Product rule: f'(x) = 2x(x³+2) + (x²-1)(3x²) = 5x⁴ - 3x² + 4x

Exercise 2: Chain Rule

Differentiate:

  1. f(x) = (3x + 1)⁵
  2. f(x) = e^(-x²)
  3. f(x) = ln(sin x)
  4. f(x) = sin²(x)
  1. f'(x) = 5(3x+1)⁴ · 3 = 15(3x+1)⁴
  2. f'(x) = e^(-x²) · (-2x) = -2xe^(-x²)
  3. f'(x) = (1/sin x) · cos x = cot x
  4. f'(x) = 2sin(x) · cos(x) = sin(2x)

Exercise 3: Combined Rules

Differentiate:

  1. f(x) = x²e^(3x)
  2. f(x) = x/√(1 + x²)
  1. Product + chain: f'(x) = 2x·e^(3x) + x²·3e^(3x) = xe^(3x)(2 + 3x)
  2. Quotient + chain: f'(x) = 1/(1+x²)^(3/2)

Exercise 4: Michaelis-Menten

For v = Vmax[S]/(Km + [S]):

  1. Find dv/d[S]
  2. Evaluate at [S] = Km
  3. What happens as [S] → ∞?
  1. dv/d[S] = VmaxKm/(Km + [S])²
  2. At [S] = Km: dv/d[S] = Vmax/(4Km)
  3. As [S] → ∞: dv/d[S] → 0 (saturation)

Coming Up: Lecture 14

Finding Extrema — Where are the maxima and minima?

The first and second derivative tests. Optimization in chemistry.

References: McQuarrie, Mathematics for Physical Chemistry; Stewart, Calculus