MTH1020 · Analysis of Change
Differentiation: Limits, Continuity & Rules
Week 7 of Monash MTH1020 Analysis of Change opens the calculus core with the derivative. It builds from limits and continuity to the limit definition f′(a) = lim(h→0)[f(a+h) − f(a)]/h, distinguishes differentiability from continuity, and assembles the toolkit: the power, product, quotient and chain rules, the standard-derivative table (exp/log/trig/inverse-trig), and implicit and logarithmic differentiation. Differentiation is examined heavily in the mid-semester tests and the final, and — because it is a foundation for Weeks 8–9 — accuracy on every rule matters.
What this chapter covers
- 01The derivative as the slope of the tangent: f′(a) = lim(h→0)[f(a+h) − f(a)]/h; Newton and Leibniz notation
- 02Limits: two-sided limit exists only if both one-sided limits agree; lim f(x) need not equal f(a)
- 03Continuity: f continuous at a if lim(x→a) f(x) = f(a); continuous classes and combinations
- 04Differentiable ⇒ continuous, but not conversely (corners |x|, vertical tangents √x, jumps)
- 05The rules: constant multiple, sum, product (fg)′ = f′g + fg′, quotient, chain (f∘g)′ = f′(g)·g′
- 06Standard derivatives: xⁿ → n x^(n−1), eˣ → eˣ, sin x → cos x, cos x → −sin x, tan x → sec²x, ln x → 1/x
- 07Inverse-trig derivatives: Sin⁻¹x → 1/√(1−x²), Cos⁻¹x → −1/√(1−x²), Tan⁻¹x → 1/(1+x²)
- 08Implicit differentiation (treat y as y(x)) and logarithmic differentiation (ln both sides) for products/powers
Product-plus-chain rule and implicit differentiation
- +1(a) Identify the structure. y = x² sin(3x) is a product of u = x² and v = sin(3x). The product rule gives y′ = u′v + uv′, so u′ = 2x, and v′ needs the chain rule.
- +1(a) Differentiate sin(3x) by the chain rule and assemble. d/dx[sin(3x)] = cos(3x) × 3 = 3cos(3x). Therefore y′ = 2x·sin(3x) + x²·3cos(3x) = 2x sin(3x) + 3x² cos(3x).
- +1(b) Differentiate the relation term by term with respect to x. From x² + y² = 25, d/dx[x²] = 2x, and d/dx[y²] = 2y·(dy/dx) by the chain rule (y is a function of x); the constant differentiates to 0. So 2x + 2y(dy/dx) = 0.
- +1(b) Solve for dy/dx. 2y(dy/dx) = −2x, so dy/dx = −x/y (for y ≠ 0). Geometrically the tangent to the circle at (x, y) is perpendicular to the radius, which −x/y encodes.
Key terms
- Derivative (f′(a))
- The instantaneous rate of change / tangent slope, f′(a) = lim(h→0)[f(a+h) − f(a)]/h, when the limit exists. Written f′(x), dy/dx or d/dx f(x).
- Continuity
- f is continuous at a if lim(x→a) f(x) = f(a). Polynomials, exponentials, logs and the trig functions are continuous on their domains; sums, products and compositions preserve continuity.
- Differentiable ⇒ continuous
- Every differentiable function is continuous, but continuity does not imply differentiability — corners (|x|), jumps and vertical tangents (√x at 0) are continuous yet non-differentiable.
- Product rule
- (fg)′ = f′g + fg′. The quotient rule is (f/g)′ = (f′g − fg′)/g² for g ≠ 0.
- Chain rule
- (f∘g)′(x) = f′(g(x))·g′(x); in Leibniz form dy/dx = (dy/du)(du/dx). The inner derivative g′ is the factor most often dropped.
- Implicit differentiation
- Differentiating a relation in x and y with respect to x, treating y as a function of x (so d/dx[y²] = 2y·dy/dx), then solving for dy/dx.
Differentiation: Limits, Continuity & Rules FAQ
Does continuous mean the same as differentiable?
No — differentiable is the stronger condition. Every differentiable function is continuous, but not every continuous function is differentiable: |x| is continuous everywhere yet has no derivative at 0 because of the sharp corner, and √x has a vertical tangent at 0. So a graph can be unbroken but still fail to have a well-defined tangent slope at certain points. Getting this implication the right way round is a common exam question.
When do I need the chain rule?
Whenever you differentiate a composition — a function of a function, like sin(3x), e^(x²) or (2x + 1)⁵. You differentiate the outer function, keep the inner function unchanged, then multiply by the derivative of the inner: d/dx sin(3x) = cos(3x)·3. Dropping that inner factor (the ×3 here) is the single most common marks-losing slip in Week-7 differentiation.
How does implicit differentiation work?
When y is tangled up with x in a relation like x² + y² = 25, you differentiate both sides with respect to x, but treat y as an unknown function of x. That means any y-term needs the chain rule: d/dx[y²] = 2y·(dy/dx). You then collect the dy/dx terms and solve algebraically, getting the derivative in terms of both x and y — here dy/dx = −x/y. It is the standard route to the derivatives of inverse functions too.
Can Sia help me with differentiation in MTH1020?
Yes. Sia can walk you through which rule a given function needs, check that you have not dropped a chain-rule factor, and take an implicit or logarithmic differentiation step by step — in the Monash write-up style. It explains the method and checks your working on your own practice questions; it does not do graded assessment for you, and Monash academic-integrity rules apply.
Exam move
Differentiation is a toolkit, so build fluency by rule. Memorise the standard-derivative table cold (including the inverse-trig entries), then drill mixed problems that force you to choose between power, product, quotient and chain — and where two apply, do them in the right order. Make the chain-rule inner factor a deliberate checkpoint on every composition. Practise implicit differentiation until the reflex 'd/dx[y²] = 2y·dy/dx' is automatic, and add logarithmic differentiation for products, quotients and variable exponents like xˣ. This chapter is heavily examined in the mid-semester tests and the final, and it is the foundation for the Week-8 and Week-9 applications, so over-learn it. Ask Sia to generate fresh derivatives across all the rules and check each line.
Working through Differentiation: Limits, Continuity & Rules in MTH1020? Sia is AskSia’s AI Mathematics tutor — ask any MTH1020 Differentiation: Limits, Continuity & Rules question and get a clear, step-by-step explanation grounded in how MTH1020 is taught and assessed. Read this chapter free, then take your hardest questions to Sia.
