# Lecture 5: Differentiation Rules — Power, Product, Quotient
## Calculus I — Differential Calculus
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## Learning Objectives
By the end of this lecture, students will be able to:
1. Apply the constant rule, constant multiple rule, and sum/difference rules
2. Use the power rule to differentiate polynomial and power functions
3. Apply the product rule to differentiate products of functions
4. Apply the quotient rule to differentiate quotients of functions
5. Combine these rules to differentiate complex expressions efficiently
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## Lecture Content
### I. Basic Differentiation Rules
The simplest differentiation rule is the **Constant Rule**: if f(x) = c is a constant, then f'(x) = 0. This makes geometric sense because the graph of a constant function is a horizontal line, which has zero slope everywhere. The **Constant Multiple Rule** states that d/dx [c * f(x)] = c * f'(x), meaning constants "factor out" of the derivative. The **Sum Rule** gives d/dx [f(x) + g(x)] = f'(x) + g'(x), and the **Difference Rule** gives d/dx [f(x) - g(x)] = f'(x) - g'(x). Taken together, these rules mean that differentiation is a **linear operator**: it respects addition and scalar multiplication.
### II. The Power Rule
The **Power Rule** states that if f(x) = x^n, then f'(x) = n * x^{n-1}. This rule is valid for any real number n, whether positive, negative, or fractional. For example, d/dx [x^5] = 5x^4, d/dx [x^{-2}] = -2x^{-3} = -2/x^3, d/dx [x^{1/2}] = (1/2)x^{-1/2} = 1/(2*sqrt(x)), and d/dx [x^{3/4}] = (3/4)x^{-1/4}.
A proof sketch for positive integers uses the limit definition: lim_{h -> 0} [(x+h)^n - x^n]/h = n*x^{n-1}, which can be established using the binomial theorem. A practical tip is to rewrite expressions involving roots and reciprocals as powers before differentiating. For instance, sqrt(x) = x^{1/2} so its derivative is (1/2)x^{-1/2}, and 1/x^3 = x^{-3} so its derivative is -3x^{-4}.
### III. Differentiating Polynomials
Since differentiation is linear, a polynomial p(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0 can be differentiated term by term, yielding p'(x) = n*a_n x^{n-1} + (n-1)*a_{n-1} x^{n-2} + ... + a_1. For example, if f(x) = 3x^4 - 5x^3 + 2x - 7, then f'(x) = 12x^3 - 15x^2 + 2.
This leads naturally to **higher-order derivatives**. The second derivative f''(x) = d/dx [f'(x)] measures the rate of change of the rate of change -- physically, if position gives velocity upon differentiation, then differentiating again gives acceleration. The notation d^2y/dx^2 is used for the second derivative, d^3y/dx^3 for the third, and so on. For a polynomial of degree n, the (n+1)th derivative is always zero.
A side-by-side comparison of f(x) = x^3 - 3x and its derivative f'(x) = 3x^2 - 3. Panel A shows the cubic curve with its local maximum and minimum marked. Panel B shows the parabola f'(x) = 3x^2 - 3, with the x-intercepts (where f'(x) = 0) aligned vertically with the extrema of f(x). Annotations connect the zero-slope points of f to the zeros of f'. Title: "A function and its derivative."
### IV. The Product Rule
The **Product Rule** states that if f and g are both differentiable, then d/dx [f(x) * g(x)] = f'(x) * g(x) + f(x) * g'(x). A helpful mnemonic is "the derivative of the first times the second, plus the first times the derivative of the second." It is essential to note that the derivative of a product is **not** the product of the derivatives; d/dx [f(x) * g(x)] does not equal f'(x) * g'(x) in general.
As an example, consider h(x) = (x^2 + 1)(x^3 - 2x). Applying the product rule gives h'(x) = 2x(x^3 - 2x) + (x^2 + 1)(3x^2 - 2) = 2x^4 - 4x^2 + 3x^4 - 2x^2 + 3x^2 - 2 = 5x^4 - 3x^2 - 2. This can be verified by expanding h(x) first to get x^5 - x^3 - 2x, whose derivative is 5x^4 - 3x^2 - 2, confirming the result.
### V. The Quotient Rule
The **Quotient Rule** states that if f and g are both differentiable and g(x) is not zero, then d/dx [f(x)/g(x)] = [f'(x) * g(x) - f(x) * g'(x)] / [g(x)]^2. A popular mnemonic is "low d-high minus high d-low, over the square of what's below."
For example, to differentiate h(x) = (x^2 + 1)/(x^3 - 1), apply the quotient rule: h'(x) = [2x(x^3 - 1) - (x^2 + 1)(3x^2)] / (x^3 - 1)^2 = [2x^4 - 2x - 3x^4 - 3x^2] / (x^3 - 1)^2 = [-x^4 - 3x^2 - 2x] / (x^3 - 1)^2. A useful tip: if the denominator is simply a constant, it is simpler to use the constant multiple rule instead of the full quotient rule.
A summary reference card showing the three main differentiation rules in a clean table format. Row 1: "Product Rule" with formula d/dx[f*g] = f'g + fg' and a small example. Row 2: "Quotient Rule" with formula d/dx[f/g] = (f'g - fg')/g^2 and a small example. Row 3: "Power Rule" with formula d/dx[x^n] = nx^{n-1} and examples for n = 3, n = -1, n = 1/2. Each row has color-coded terms to help identify f, g, f', g' in the formulas.
### VI. Combining Rules — Worked Examples
Most real-world functions require combining multiple differentiation rules. Consider y = x^2 * sqrt(x) = x^{5/2}. The power rule gives y' = (5/2)x^{3/2} directly, though one could also arrive at the same result by applying the product rule to x^2 * x^{1/2}.
As another example, y = (3x + 1)/(x^2 + 4) requires the quotient rule: y' = [3(x^2 + 4) - (3x + 1)(2x)] / (x^2 + 4)^2 = [3x^2 + 12 - 6x^2 - 2x] / (x^2 + 4)^2 = [-3x^2 - 2x + 12] / (x^2 + 4)^2. A general strategy is to simplify the function first if possible -- dividing, factoring, or canceling -- before differentiating, since simpler expressions lead to fewer errors.
### VII. Higher-Order Derivatives
The second derivative f''(x) = d/dx [f'(x)] carries important physical meaning: if s(t) represents position, then s'(t) is velocity and s''(t) is acceleration. The nth derivative f^{(n)}(x) is obtained by differentiating n times. For polynomials of degree n, the (n+1)th derivative is always zero, since each differentiation reduces the degree by one.
To illustrate, consider f(x) = x^4 - 2x^3 + x. The successive derivatives are f'(x) = 4x^3 - 6x^2 + 1, f''(x) = 12x^2 - 12x, f'''(x) = 24x - 12, f^{(4)}(x) = 24, and f^{(5)}(x) = 0. From the fourth derivative onward, all higher derivatives vanish.