Unit 3: Differential calculus
Mathematics-II notes · PTU syllabus (BSIT202/BSBC202)
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Unit summary
Differentiation measures rates of change and finds maximum and minimum values. This unit covers an introduction to differentiation, the derivative of a function of one variable, power functions, sum and product of two functions, function of a function, differentiation by substitution, and maxima and minima.
After this unit you can
- Explain the derivative as a rate of change
- Differentiate power, sum, product and quotient functions
- Apply the chain rule and substitution
- Find maxima and minima
PTU syllabus topics
- Introduction to differentiation
- derivative of a function of one variable
- power functions
- sum and product of two functions
- function of a function
- differentiation by substitution
- maxima and minima
Power rule
d/dx (xⁿ) = n xⁿ⁻¹
Product rule
(uv)' = u'v + uv'
Quotient rule
(u/v)' = (u'v − uv') / v²
Chain rule
dy/dx = dy/du × du/dx
Maxima and minima
f'(x) = 0; f'' < 0 max, f'' > 0 min
Topic 1
Introduction to differentiation
- Derivative: the instantaneous rate of change of y with respect to x — the slope of the tangent to the curve.
First principles
dy/dx = lim (h → 0) [f(x + h) − f(x)] ÷ h
Topic 2
Standard derivatives
Power rule
d/dx (xⁿ) = n xⁿ⁻¹
Constant
d/dx (c) = 0
Exponential
d/dx (eˣ) = eˣ; d/dx (aˣ) = aˣ log a
Logarithm
d/dx (log x) = 1/x
Trigonometric
d/dx (sin x) = cos x; d/dx (cos x) = −sin x
Topic 3
Sum, product and quotient rules
Sum
(u + v)′ = u′ + v′
Product
(uv)′ = u v′ + v u′
Quotient
(u/v)′ = (v u′ − u v′) ÷ v²
Example
y = (x² + 1)(3x − 2): dy/dx = (x² + 1)(3) + (3x − 2)(2x) = 9x² − 4x + 3.
Topic 4
Function of a function and differentiation by substitution
Rule
If y = f(u) and u = g(x), dy/dx = (dy/du) × (du/dx)
Example
y = (3x² + 5)⁴: let u = 3x² + 5; dy/dx = 4u³ × 6x = 24x(3x² + 5)³.
- Substitution: replace an inner expression by u, differentiate, then substitute back.
Topic 5
Maxima and minima
- 1Find dy/dx and set it to 0
- 2Solve for critical points
- 3Find d²y/dx² at each point
- 4Negative → maximum; positive → minimum
Example
Profit P = −2x² + 40x − 50: dP/dx = −4x + 40 = 0 → x = 10; d²P/dx² = −4 < 0, so maximum profit = −200 + 400 − 50 = 150.
- Business uses: maximising profit and revenue, minimising cost; marginal cost and marginal revenue are derivatives.
Key terms
- Derivative
- Rate of change of a function
- Chain rule
- Rule for differentiating a function of a function
- Critical point
- Point where the derivative is zero
- Second derivative test
- Using d²y/dx² to classify extrema
- Marginal cost
- Derivative of total cost
Quick revision
- First principles definition.
- Power, exponential, log, trig derivatives.
- Sum, product, quotient rules.
- Chain rule and substitution.
- Maxima and minima by first and second derivatives; marginal concepts.
Important exam questions
Practice questions written to the PTU exam pattern for this unit's syllabus: short answers (Section A style) and long answers (Sections B and C style).
Short-answer questions
- Q1.Define the derivative.
- Q2.Differentiate x⁵ + 3x².
- Q3.State the product rule.
- Q4.Differentiate (2x + 1)³.
- Q5.State the condition for a maximum.
- Q6.What is marginal revenue?
Long-answer questions
- Q1.Differentiate functions using sum, product and quotient rules (numerical).
- Q2.Explain the chain rule with examples.
- Q3.Find maxima and minima of a function (numerical).
- Q4.Apply differentiation to a profit maximisation problem (numerical).
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