Premed · Premed · Calculus 1
Lecture 5: Differentiation Rules — Power, Product, Quotient
Calculus I — Differential Calculus
Learning Objectives
By the end of this lecture, students will be able to:
- Apply the constant rule, constant multiple rule, and sum/difference rules
- Use the power rule to differentiate polynomial and power functions
- Apply the product rule to differentiate products of functions
- Apply the quotient rule to differentiate quotients of functions
- Combine these rules to differentiate complex expressions efficiently
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/(2sqrt(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) = na_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.
<image>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."</image>
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.
<image>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.</image>
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.
