Difficulty: Intermediate | Prerequisites: Limits and Average Rate of Change notes
The derivative is the single most important concept in MAT 210. It tells you the instantaneous rate at which a function is changing at any given point, and every application topic in the course (marginal analysis, optimisation, graph sketching) depends on your ability to compute derivatives quickly and accurately. This section covers what a derivative is, the rules for finding one efficiently, and how those rules apply to the algebraic, exponential, and logarithmic functions listed in the syllabus.
The derivative of a function at a point is the limit of the difference quotient as the interval shrinks to zero, giving you the instantaneous rate of change rather than the average. A set of rules (power, product, quotient, chain) lets you compute derivatives of nearly any function you will meet in this course without going back to the limit definition each time.
Derivative
The function f'(x) = lim (h→0) [f(x + h) - f(x)] / h. It gives the slope of the tangent line to the graph of f at any point x. Think of it as the speedometer reading at one exact moment, as opposed to your average speed over a trip.
Instantaneous rate of change
Another name for the derivative at a specific point. The rate at which f is changing right at x = c, not over an interval.
Differentiable
A function is differentiable at x = c if f'(c) exists, meaning the limit definition produces a finite number. In simple terms, the graph is smooth at that point, with no sharp corner, cusp, or vertical tangent.
Tangent line
The line that just touches the curve at a single point and has slope equal to the derivative there. Its equation is y - f(a) = f'(a)(x - a).
Power rule
If f(x) = xⁿ, then f'(x) = nxⁿ⁻¹. The exponent comes down as a coefficient, and the exponent decreases by one.
Product rule
For f(x) = u(x) · v(x), the derivative is f'(x) = u'(x)v(x) + u(x)v'(x). Think of it as: "first times derivative of second, plus second times derivative of first."
Quotient rule
For f(x) = u(x)/v(x), the derivative is f'(x) = [u'(x)v(x) - u(x)v'(x)] / [v(x)]². The mnemonic "low d-high minus high d-low, over the square of what's below" captures the pattern.
Chain rule
For a composite function f(g(x)), the derivative is f'(g(x)) · g'(x). Differentiate the outer function (keeping the inner function untouched), then multiply by the derivative of the inner function.
Exponential derivative
The derivative of eˣ is eˣ. For the general exponential aˣ, the derivative is aˣ · ln(a).
Logarithmic derivative
The derivative of ln(x) is 1/x. For log base a, the derivative of log_a(x) is 1/(x · ln(a)).
f'(x) = lim (h→0) [f(x + h) - f(x)] / h.
At a specific point x = a, f'(a) gives the slope of the tangent line at that point.
Example: for f(x) = x², the difference quotient simplifies to 2x + h, so f'(x) = lim (h→0) (2x + h) = 2x.
You will not use the limit definition for every derivative. The rules below are shortcuts derived from it.
If f(x) = xⁿ, then f'(x) = nxⁿ⁻¹. Works for any real exponent, including negatives and fractions.
The derivative of a constant is 0 (a constant is x⁰, and 0 · x⁻¹ = 0).
Example: f(x) = x⁷ gives f'(x) = 7x⁶.
Example: f(x) = 1/x = x⁻¹ gives f'(x) = -x⁻² = -1/x².
Example: f(x) = √x = x^(1/2) gives f'(x) = (1/2)x^(-1/2) = 1/(2√x).
Constant multiple: d/dx [c · f(x)] = c · f'(x). Constants pass through the derivative.
Sum/difference: d/dx [f(x) ± g(x)] = f'(x) ± g'(x). Differentiate term by term.
Example: d/dx (5x³ - 2x + 7) = 15x² - 2.
For y = u · v: y' = u'v + uv'.
Example: y = (3x²)(eˣ). Then u = 3x², u' = 6x, v = eˣ, v' = eˣ. So y' = 6xeˣ + 3x²eˣ = (6x + 3x²)eˣ.
Common error: students multiply the two derivatives together. That is wrong. The product rule has two terms added, not multiplied.
For y = u/v: y' = (u'v - uv') / v².
Example: y = (x² + 1)/(x - 3). u = x² + 1, u' = 2x, v = x - 3, v' = 1. So y' = [2x(x - 3) - (x² + 1)(1)] / (x - 3)² = (2x² - 6x - x² - 1) / (x - 3)² = (x² - 6x - 1) / (x - 3)².
Tip: if the denominator is just a constant, use the constant multiple rule instead. No need for the quotient rule on f(x)/5.
For y = f(g(x)): y' = f'(g(x)) · g'(x). "Derivative of the outside, times derivative of the inside."
Example: y = (2x + 1)⁵. The outer function is u⁵, the inner function is 2x + 1. So y' = 5(2x + 1)⁴ · 2 = 10(2x + 1)⁴.
Example: y = e^(3x²). y' = e^(3x²) · 6x = 6xe^(3x²).
Example: y = ln(x² + 4). y' = [1/(x² + 4)] · 2x = 2x/(x² + 4).
The chain rule is the rule that trips up the most students. Whenever you see a function inside another function, you need it.
d/dx (eˣ) = eˣ. The exponential function is its own derivative.
d/dx (aˣ) = aˣ · ln(a), where a > 0 and a ≠ 1.
d/dx (ln x) = 1/x, for x > 0.
d/dx (log_a x) = 1/(x · ln a).
Combined with the chain rule: d/dx [e^(g(x))] = e^(g(x)) · g'(x), and d/dx [ln(g(x))] = g'(x)/g(x).
The second derivative f''(x) is the derivative of f'(x). It measures how the rate of change itself is changing.
In applications: if f(x) is position, f'(x) is velocity, and f''(x) is acceleration.
You may see the notation f''(x), d²y/dx², or y''.
f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\frac{d}{dx}[x^n] = nx^{n-1}\frac{d}{dx}[u \cdot v] = u'v + uv'\frac{d}{dx}\left[\frac{u}{v}\right] = \frac{u'v - uv'}{v^2}\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\frac{d}{dx}[e^x] = e^x \qquad \frac{d}{dx}[a^x] = a^x \ln a\frac{d}{dx}[\ln x] = \frac{1}{x} \qquad \frac{d}{dx}[\log_a x] = \frac{1}{x \ln a}Tangent line at x = a:
y - f(a) = f'(a)(x - a)The derivative is the mathematical engine behind any question of the form "how fast is this changing right now?" Economists use derivatives to find marginal cost and marginal revenue (the cost or revenue from producing one additional unit). Biologists use them to model the rate of bacterial growth at a given moment. Engineers use them to calculate the velocity of a moving object at a precise instant.
Multiplying two derivatives together instead of applying the product rule. The derivative of u · v is u'v + uv', not u' · v'.
Forgetting the chain rule when a function is composed. d/dx [e^(3x)] is not just e^(3x); it is 3e^(3x).
Treating d/dx (ln 5) as 1/5. ln 5 is a constant, so its derivative is 0.
Confusing d/dx [xⁿ] = nxⁿ⁻¹ with d/dx [aˣ] = aˣ ln a. The first has the variable in the base; the second has the variable in the exponent. They are completely different rules.
⚠️ The midterm and final both test every differentiation rule. Speed and accuracy matter, so drill these until they are reflexive.
⚠️ Chain rule problems involving eˣ and ln(x) composites appear heavily. Expect expressions like e^(2x+1) or ln(x³ + 5).
⚠️ Tangent line equations appear on exams. You need the derivative evaluated at the point, plus the point-slope formula.
True or false: the derivative of a constant function is 1. Answer: False. The derivative of any constant is 0.
Fill in the blank: the product rule says d/dx [u · v] = ______ + ______. Answer: u'v + uv'.
True or false: d/dx [ln(3x)] = 1/(3x). Answer: False. By the chain rule, d/dx [ln(3x)] = (1/(3x)) · 3 = 1/x.
What is d/dx [e^(5x)]? Answer: 5e^(5x).
What is the derivative of f(x) = x⁴ - 3x² + 2? Answer: f'(x) = 4x³ - 6x.
Q: Find f'(x) if f(x) = (x³ + 2x)(4x - 1).
A: Product rule. u = x³ + 2x, u' = 3x² + 2, v = 4x - 1, v' = 4. f'(x) = (3x² + 2)(4x - 1) + (x³ + 2x)(4) = 12x³ - 3x² + 8x - 2 + 4x³ + 8x = 16x³ - 3x² + 16x - 2.
Q: Find the derivative of y = (2x - 5)⁴.
A: Chain rule. y' = 4(2x - 5)³ · 2 = 8(2x - 5)³.
Q: Find dy/dx for y = ln(x² + 1).
A: Chain rule on ln. dy/dx = [1/(x² + 1)] · 2x = 2x/(x² + 1).
Q: Write the equation of the tangent line to f(x) = x³ at x = 2.
A: f(2) = 8. f'(x) = 3x², so f'(2) = 12. Tangent line: y - 8 = 12(x - 2), or y = 12x - 16.
Q: Find d/dx [5^x].
A: d/dx [aˣ] = aˣ ln a. So d/dx [5^x] = 5^x · ln 5.
The differentiation rules here are prerequisites for everything that follows in MAT 210. Marginal analysis applies derivatives to cost, revenue, and profit functions. Optimisation uses derivatives set to zero to find maximum and minimum values. Graph analysis uses the first and second derivatives to determine where a function increases, decreases, and changes concavity.
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