🎓 UP FIRST-YEAR STEM · SEMESTER 1 2026

The structured support your module deserves

60 minutes. Postgraduate facilitator. Small groups. Built around the exact problems in your UP module right now — not generic content.

Join free — first session on us → See my module ↓

67%

of WTW 114 students fail Sem Test 1

R100

per session — R800/month

Free

First session — no payment needed

Six modules · Semester 1 2026

Your module. Your facilitator. Your session.

Every session is built around the specific pain points in your UP module right now — not a generic maths tutorial.

WTW 114 ·Calculus — NAS & EBiT

Join group → Tue · Thu 19:00

WTW 124 ·Calculus 2 — NAS

Tue · Thu 19:00

WTW 158 ·Engineering Mathematics — EBiT

Tue · Thu 19:00

WTW 164 ·Engineering Mathematics 2 — EBiT

Tue · Thu 19:00

STK 110 ·Statistics — EMS & NAS

Tue · Thu 19:00

WST 111 ·Mathematical Statistics — NAS

Tue · Thu 19:00

Four phases · 60 minutes

Every session follows the same structure

This is not a lecture. It is not a recording you can pause and ignore. It is 60 minutes of structured, expert-led practice — designed around the exam problems you will actually face.

Ambush — 10 min

A cold problem. No hints. No preview.

Dropped to all students simultaneously. The discomfort is intentional — it shows exactly where your understanding breaks.

Anatomy — 15 min

Your facilitator dissects the approach.

Step by step. Socratic questions at every decision point. Not a solution walkthrough — a guided reconstruction of the correct thinking.

Drill — 25 min

Entry · Standard · Stretch. Leaderboard live.

Three problems calibrated to three difficulty levels. Six minutes each. Your facilitator reads the room in real time.

Close — 10 min

One principle. One exit ticket.

The session encoded into a single sentence you can put on a flash card. AI summary to your inbox within 30 minutes.

Pricing

April 2026 · Semester 1

R800

per month · 8 sessions · April intake

R100 per session — same rate across all 8 sessions

Tuesday and Thursday evenings · 19:00–20:00
Postgraduate facilitator — same person every session
Maximum small group sessions
Session recording available after every session
AI summary to your email within 30 minutes
First session completely free

Re-registering a failed module costs R7,500+. One month of UMM costs R800. The maths is straightforward.

Join now — first session free →

Top-Up Sessions

Need an extra session? R100 per Top-Up.

Available for enrolled students. Saturday morning slots. Exam-prep intensives. Book via the Top-Up channel in the WhatsApp Community.

Book a Top-Up →

Ready?

Join the Community. First session free.

Reply with your name and module. We will confirm your session time and send your UMM platform join link before Tuesday.

WhatsApp us to enrol →

+27 61 009 5548 · uni.maths.mastery@unimathsmastery.com

Before you commit

Questions students actually ask

Is the first session really free? No payment needed?

Correct. Your first session requires no payment and no commitment. Attend, experience the structure, meet your facilitator. If you decide UMM is for you — pay R800 for the month. If not — no obligation. We would rather you experience the product than just read about it.

What if I miss a session?

Sessions are recorded automatically on the UMM platform. If you miss a Tuesday session, the recording is available in your dashboard before Thursday. You do not lose money for a missed session — your R800 covers the month, not individual attendance. Consistent attendance produces results; we track it and so do your parents, but missing one session is not a crisis.

How is this different from a UP tutorial?

UP tutorials typically cover problem sets with a tutor who has limited time and 20+ students waiting. UMM sessions follow a fixed four-phase structure built around the specific pain points in your module right now — not the prescribed problem set. Your facilitator is a postgraduate specialist, not a fellow student. The session is designed to build exam-condition problem-solving, not homework completion.

Can I join mid-semester?

Yes. The April intake runs until 22 May 2026. If you join after the first session, your UMM platform access gives you the recordings of every session you missed. WhatsApp us and we will confirm your start date and pro-rate accordingly if appropriate.

What do I need to attend a session?

A device with a stable internet connection — phone, tablet, or laptop all work. Earphones are strongly recommended. A pen and paper for the Ambush and Drill problems. Your UMM platform login, which is set up when you enrol. That is everything.

Can I attend sessions for more than one module?

Yes — if you are enrolled in two of the six modules and want sessions for both, WhatsApp us. Multi-module enrolment is handled individually and pricing is discussed based on your specific combination.

Do I need to pay before the first session?

No. Your first session is free. Payment of R800 is required to continue after the first session and to access the full month — 8 sessions, recordings, AI summaries, and your module sub-group in the WhatsApp Community.

My module is not listed. Can I still join?

Not yet for Semester 1 2026. WhatsApp us with your module code and we will add you to the waitlist for Semester 2 expansion. UMM is actively expanding the module list based on demand.

Resource library · Semester 1 2026

Your module. Right now. UMM style.

Every card below covers a topic from your current UP semester — matched to the module calendar and study guide. Each card shows the Ambush problem, how UMM's facilitator approaches it, and a Drill problem at standard level. Solutions are worked through live in the session.

⚡ NEXT SESSION = topics covered in the very first session — Thursday 3 April 19:00 on the UMM platform

WTW 114 · The Chain Rule

Week 5 · Pre-Test 1

⚡ Ambush — solve this cold before reading further

Differentiate h(x) = sin(x² + 3x). Identify the outer and inner function explicitly before differentiating.

🔬 UMM Anatomy — how to approach this

Chain Rule (Study Guide Unit 3.5): d/dx[f(g(x))] = f′(g(x))·g′(x). Outer function f(u) = sin u, inner g(x) = x² + 3x.
The UP formula sheet lists: d/dx[sin f(x)] = cos(f(x))·f′(x). Apply directly: h′(x) = cos(x² + 3x)·(2x + 3).
Extended Chain Rule for three layers — work strictly outside-in, one layer at a time. Never skip layers.
Most common error: computing d/dx[sin(x²+3x)] = cos(x²+3x) and stopping. The ·(2x+3) is non-negotiable.

🎯 Drill — standard level

Find dy/dx if y = (3x² − 2x + 1)⁵. Use the Chain Rule form d/dx[f(x)]ⁿ = n[f(x)]ⁿ⁻¹·f′(x) from your study guide.

Full solution worked through live in the session →

Tue & Thu · 19:00–20:00 · UMM platform Join — first session free →

WTW 114 · Implicit Differentiation

Week 5/7 · In Test 1 scope

⚡ Ambush — solve this cold before reading further

Find dy/dx given x³ + y³ = 6xy. The study guide (Unit 3.8) uses the same approach for x² + y² = 25. Do not solve for y explicitly.

🔬 UMM Anatomy — how to approach this

The Implicit Differentiation theorem guarantees y = f(x) is differentiable — differentiate both sides w.r.t. x simultaneously.
Every y-term requires the Chain Rule: d/dx[y³] = 3y²·(dy/dx). Write y′ throughout for clarity, as the study guide does.
Right side needs the Product Rule: d/dx[6xy] = 6·(dx/dx)·y + 6x·(dy/dx) = 6y + 6x·y′.
After differentiating: 3x² + 3y²y′ = 6y + 6xy′. Collect y′ terms: y′(3y² − 6x) = 6y − 3x². Factorise and isolate.

🎯 Drill — standard level

Use implicit differentiation to find y′ if x² + 2xy − y³ = 7. Express your answer in terms of x and y.

Full solution worked through live in the session →

Tue & Thu · 19:00–20:00 · UMM platform Join — first session free →

WTW 114 · Derivatives of Log & Inverse Trig

Week 7 · Last week before recess

⚡ Ambush — solve this cold before reading further

Find d/dx[arctan(x²)]. The study guide (Unit 3.7, Theorem 10) proves d/dx[arctan x] = 1/(1+x²). Apply the Chain Rule extension.

🔬 UMM Anatomy — how to approach this

From the study guide proof: d/dx[arctan x] = 1/(1+x²). With Chain Rule (Unit 3.7 item 11): d/dx[arctan(f(x))] = f′(x)/(1+(f(x))²).
For d/dx[ln(f(x))]: result is f′(x)/f(x). Always the derivative of the argument over the argument itself.
Logarithmic differentiation for y = x^x: take ln of both sides → ln y = x ln x → differentiate implicitly → y′/y = ln x + 1.
Sign distinction: d/dx[arcsin x] = 1/√(1−x²) but d/dx[arccos x] = −1/√(1−x²). The negative sign on arccos is a common mark loss.

🎯 Drill — standard level

Differentiate y = ln(x² + 1) + arctan(3x). Use the Chain Rule extensions from the study guide for both terms.

Full solution worked through live in the session →

Tue & Thu · 19:00–20:00 · UMM platform Join — first session free →

WTW 114 · Exponential Growth & Decay

Week 7 · Last week before recess

⚡ Ambush — solve this cold before reading further

A substance decays as m(t) = m₀e^(−kt). The half-life is 8 years. Find k exactly. Then find what fraction of m₀ remains after 20 years. Leave your answer in exact form.

🔬 UMM Anatomy — how to approach this

Set up the half-life condition: m(8) = m₀/2 gives m₀e^(−8k) = m₀/2. Cancel m₀: e^(−8k) = 1/2.
Take ln of both sides: −8k = ln(1/2) = −ln 2, so k = ln(2)/8. This is the standard UP result for half-life problems.
For m(20): substitute k back → m(20) = m₀·e^(−20·ln2/8) = m₀·e^(−5ln2/2) = m₀·2^(−5/2) = m₀/(4√2).
Growth problems use the identical structure with k > 0. The differentiation rules from Unit 3.6 apply throughout: d/dx[e^(f(x))] = f′(x)·e^(f(x)).

🎯 Drill — standard level

A population grows from 1000 to 2500 in 5 years. Using P(t) = P₀e^(kt), find k exactly and predict the population after 12 years.

Full solution worked through live in the session →

Tue & Thu · 19:00–20:00 · UMM platform Join — first session free →

WTW 114 · Related Rates

NEXT SESSION

⚡ Ambush — solve this cold before reading further

A spherical balloon is being inflated. When the radius is 5 cm, the radius is increasing at 0.3 cm/s. How fast is the volume increasing at that instant? (V = 4πr³/3)

🔬 UMM Anatomy — how to approach this

Step 1 — Identify all quantities changing with time. Here: r(t) and V(t). Given: dr/dt = 0.3 cm/s when r = 5. Find: dV/dt.
Step 2 — Write the geometric relationship: V = (4/3)πr³. This connects the two quantities.
Step 3 — Differentiate both sides with respect to t: dV/dt = 4πr²·(dr/dt). This is the Chain Rule applied to a time-varying function.
Step 4 — Substitute the known values at the specific instant: dV/dt = 4π(5)²(0.3) = 30π cm³/s. Always state units.

🎯 Drill — standard level

A ladder 10 m long leans against a wall. Its base slides away at 0.4 m/s. How fast is the top sliding down when the base is 6 m from the wall? (Hint: x² + y² = 100.)

Full solution worked through live in the session →

Tue 3 April · 19:00–20:00 · First session · UMM platform Join — first session free →

WST 111 · Conditional Probability

Week 4 · Tutorial Test 1 scope

⚡ Ambush — solve this cold before reading further

A factory has two machines. Machine A produces 60% of items, Machine B produces 40%. Machine A has a 3% defect rate, Machine B has a 5% defect rate. An item is selected at random and found to be defective. What is the probability it was made by Machine B?

🔬 UMM Anatomy — how to approach this

Define events using UP notation: let B = 'made by Machine B', D = 'defective'. You need P(B|D).
Apply Bayes' Theorem (Study Guide Section 2.4): P(B|D) = P(D|B)·P(B) / P(D). Identify each value from the problem.
Law of Total Probability for P(D): P(D) = P(D|A)·P(A) + P(D|B)·P(B) = (0.03)(0.60) + (0.05)(0.40) = 0.018 + 0.020 = 0.038.
P(B|D) = (0.05)(0.40)/0.038 = 0.020/0.038 ≈ 0.526. Machine B is more likely — despite producing fewer items, its higher defect rate makes it the more probable source.

🎯 Drill — standard level

Box A contains 4 red and 2 blue balls. Box B contains 1 red and 5 blue balls. A box is chosen with P(A) = 0.4. A ball is drawn and is red. Use Bayes' Theorem to find P(A|red).

Full solution worked through live in the session →

Tue & Thu · 19:00–20:00 · UMM platform Join — first session free →

WST 111 · Discrete Random Variables

Week 5 · Tutorial Test 2 scope

⚡ Ambush — solve this cold before reading further

A random variable X has pmf p(x) = cx for x = 1, 2, 3, 4, and 0 otherwise. Find the value of c, then calculate P(X ≥ 3) and F(2).

🔬 UMM Anatomy — how to approach this

Valid pmf requirement (Study Guide Section 3.2): Σp(x) = 1 over all x. Set up: c(1+2+3+4) = 10c = 1, so c = 1/10.
P(X ≥ 3) = P(X=3) + P(X=4) = 3/10 + 4/10 = 7/10. Or use complement: 1 − P(X ≤ 2) = 1 − (1/10 + 2/10) = 7/10.
CDF: F(2) = P(X ≤ 2) = p(1) + p(2) = 1/10 + 2/10 = 3/10. The CDF is non-decreasing, right-continuous, F(−∞) = 0, F(+∞) = 1.
Always verify: all p(x) ≥ 0 and Σp(x) = 1. A pmf that fails either condition is invalid regardless of the rest of the calculation.

🎯 Drill — standard level

Let X have pmf p(x) = k(3−x) for x = 0, 1, 2. Find k, the CDF F(x) for all x, and P(0.5 < X < 2.5).

Full solution worked through live in the session →

Tue & Thu · 19:00–20:00 · UMM platform Join — first session free →

WST 111 · Expected Values & Variance

Week 7 · Post-Semester Test 1

⚡ Ambush — solve this cold before reading further

X has pmf p(0) = 0.3, p(1) = 0.4, p(2) = 0.2, p(3) = 0.1. Find E(X), Var(X), and E(2X² − 3X + 1). Show all working.

🔬 UMM Anatomy — how to approach this

E(X) = Σ x·p(x) = 0(0.3) + 1(0.4) + 2(0.2) + 3(0.1) = 0 + 0.4 + 0.4 + 0.3 = 1.1.
E(X²) = Σ x²·p(x) = 0(0.3) + 1(0.4) + 4(0.2) + 9(0.1) = 0 + 0.4 + 0.8 + 0.9 = 2.1. Note: E(X²) ≠ [E(X)]².
Var(X) = E(X²) − [E(X)]² = 2.1 − (1.1)² = 2.1 − 1.21 = 0.89. This is the computing formula — always faster than the definition.
For E(2X²−3X+1): use linearity. E(2X²−3X+1) = 2E(X²) − 3E(X) + 1 = 2(2.1) − 3(1.1) + 1 = 4.2 − 3.3 + 1 = 1.9.

🎯 Drill — standard level

Let X have pmf p(x) = (4−x)/10 for x = 0, 1, 2, 3. Find E(X), Var(X), and σ. Then find Var(3X−2) using the rule Var(aX+b) = a²Var(X).

Full solution worked through live in the session →

Tue & Thu · 19:00–20:00 · UMM platform Join — first session free →

WST 111 · Moments & Moment Generating Functions

NEXT SESSION

⚡ Ambush — solve this cold before reading further

The MGF of X is M_X(t) = (0.4 + 0.6e^t)^8. Without finding the distribution table, find E(X) and Var(X) using derivatives of M_X(t).

🔬 UMM Anatomy — how to approach this

MGF method: E(X) = M′_X(0) and E(X²) = M″_X(0). Differentiate M_X(t) w.r.t. t, then evaluate at t = 0.
M′_X(t) = 8(0.4 + 0.6e^t)^7·(0.6e^t). At t = 0: M′_X(0) = 8(1)^7·(0.6) = 4.8. So E(X) = 4.8.
Recognise the structure: M_X(t) = (q + pe^t)^n is the MGF of Binomial(n=8, p=0.6). This gives E(X) = np = 8(0.6) = 4.8 directly.
Var(X) = np(1−p) = 8(0.6)(0.4) = 1.92 for the Binomial. Verifying via M″_X(0) − [M′_X(0)]² gives the same result.

🎯 Drill — standard level

Find E(X) and Var(X) if M_X(t) = e^(3t + 2t²). Identify the named distribution and state its parameters μ and σ².

Full solution worked through live in the session →

Tue 3 April · 19:00–20:00 · First session · UMM platform Join — first session free →

WST 111 · Binomial Distribution

NEXT SESSION

⚡ Ambush — solve this cold before reading further

20% of products from a production line are defective. A quality inspector randomly selects 12 items. Find: (a) P(exactly 3 defective), (b) P(fewer than 2 defective), (c) E(X) and Var(X).

🔬 UMM Anatomy — how to approach this

Check the four Binomial conditions: fixed n=12, independent trials, constant p=0.20, binary outcome (defective/not). All satisfied → X ~ B(12, 0.20).
P(X=3) = C(12,3)·(0.20)³·(0.80)⁹. Compute: C(12,3) = 220, (0.20)³ = 0.008, (0.80)⁹ ≈ 0.1342. Result: 220 × 0.008 × 0.1342 ≈ 0.236.
P(X < 2) = P(X=0) + P(X=1). P(X=0) = (0.80)¹² ≈ 0.0687. P(X=1) = 12·(0.20)·(0.80)¹¹ ≈ 0.2062. Sum ≈ 0.275.
For Binomial: E(X) = np = 12(0.20) = 2.4. Var(X) = np(1−p) = 12(0.20)(0.80) = 1.92. These formulas must be memorised.

🎯 Drill — standard level

X ~ B(n=15, p=0.35). Find: (a) P(X=5), (b) P(X ≤ 3), (c) P(X > 10), (d) E(X) and σ. Show all working using the binomial formula.

Full solution worked through live in the session →

Tue 3 April · 19:00–20:00 · First session · UMM platform Join — first session free →

Join UMM — first session free →

R800/month · 8 sessions

How it breaks down

R100

per workshop

sessions/month minimum

FREE

first session — no commitment

· Tue & Thu 19:00 · Small groups · UMM platform

WhatsApp module groups

Join your module group

Each module has its own dedicated WhatsApp group inside the UMM Community. Session reminders, topic previews, and facilitator updates — all in one place.

WTW 114 ·Calculus — NAS & EBiT

Join group →

WTW 124 ·Calculus 2 — NAS

Join group →

WTW 158 ·Engineering Maths — EBiT

Join group →

WTW 164 ·Engineering Maths 2 — EBiT

Join group →

STK 110 ·Statistics — EMS & NAS

Join group →

WST 111 ·Maths Statistics — NAS

Join group →