Basic MathematicsUnit 412 min read

Differentiation – Definitions, Rules, Applications & Real‑World Examples

Unit 4 of Basic Mathematics: introduces the concept of derivative, fundamental differentiation rules, higher‑order derivatives, and practical applications such as rates of change, optimisation and related‑rates problems.

Padhai · Based on the official syllabus and past papers
Updated 5 Oct 2026

Key points

The derivative measures instantaneous rate of change and the slope of the tangent to a curve.
Mastery of basic rules (power, product, quotient, chain) enables quick differentiation of any elementary function.
Higher‑order derivatives and implicit differentiation extend the technique to curves defined implicitly and to motion problems.
Differentiation underpins optimisation, marginal analysis in economics, and real‑time monitoring in tech services.
Exam questions typically test rule application, interpretation of derivative graphs, and problem‑solving with rates or extrema.

1. What is a Derivative?

The derivative of a function $f$ at a point $x = a$ is the limit

Tangent line to y = x³ − 3x² + 2 at x=1 (slope = f'(1) = -1)

Tangent line to y = x² at x=1 (slope = f'(1) = 2)

$f^{'} (a) = h \to 0 lim \frac{f ( a + h ) - f ( a )}{h}$

If the limit exists, $f$ is differentiable at $a$ . Geometrically, $f^{'} (a)$ is the slope of the tangent line to the curve $y = f (x)$ at $(a, f (a))$ .

Visual – Tangent Slope

2. Basic Differentiation Rules

Rule	Formula	When to Use
Power Rule	$\frac{d}{d x} x^{n} = n x^{n - 1}$	Any monomial $x^{n}$ (n any real)
Constant Multiple	$\frac{d}{d x} [c f (x)] = c f^{'} (x)$	Constant factor outside
Sum/Difference	$\frac{d}{d x} [f \pm g] = f^{'} \pm g^{'}$	Linear combination
Product Rule	$\frac{d}{d x} [f g] = f^{'} g + f g^{'}$	Product of two functions
Quotient Rule	$\frac{d}{d x} [\frac{f}{g}] = \frac{f ^{'} g - f g ^{'}}{g ^{2}}$	Ratio of two functions
Chain Rule	$\frac{d}{d x} f (g (x)) = f^{'} (g (x)) \cdot g^{'} (x)$	Composite functions

Critical points of f(x) = x³ − 3x² + 2 (where f'(x) = 0 or undefined)

Worked Example 1 – Applying Multiple Rules

Find $\frac{d}{d x} [(3 x^{2} + 2) sin (x^{3})]$ .

Solution

Identify outer product: $u (x) = 3 x^{2} + 2$ , $v (x) = sin (x^{3})$ .
Apply Product Rule: $(uv)^{'} = u^{'} v + u v^{'}$ .
- $u^{'} = \frac{d}{d x} (3 x^{2} + 2) = 6 x$ (Power Rule).
- $v^{'} = \frac{d}{d x} sin (x^{3})$ . Use Chain Rule:
  
  $v^{'} = cos (x^{3}) \cdot \frac{d}{d x} (x^{3}) = cos (x^{3}) \cdot 3 x^{2} .$
Assemble:

$\frac{d}{d x} [(3 x^{2} + 2) sin (x^{3})] = (6 x) sin (x^{3}) + (3 x^{2} + 2) [3 x^{2} cos (x^{3})] = 6 x sin (x^{3}) + 3 x^{2} (3 x^{2} + 2) cos (x^{3}) .$

Result: $6 x sin (x^{3}) + 3 x^{2} (3 x^{2} + 2) cos (x^{3})$ .

Visual – Graph of the Result

Derivative (green) of f(x) = (3x²+2)·sin(x³) vs. original function (dashed blue)

3. Differentiation of Special Functions

Function Type	Derivative
Exponential $e^{x}$	$e^{x}$
General exponential $a^{x}$	$a^{x} ln a$
Natural log $ln x$	$\frac{1}{x}$
Log base $a$ $lo g_{a} x$	$\frac{1}{x l n a}$
Trigonometric $sin x$	$cos x$
$cos x$	$- sin x$
$tan x$	$sec^{2} x$
$cot x$	$- csc^{2} x$
$sec x$	$sec x tan x$
$csc x$	$- csc x cot x$

Worked Example 2 – Log‑Exponential Combination

Differentiate $y = x^{2} e^{3 x}$ .

Solution

Treat as product: $u = x^{2}$ , $v = e^{3 x}$ .
$u^{'} = 2 x$ .
$v^{'} = e^{3 x} \cdot 3$ (Chain Rule).

$y^{'} = u^{'} v + u v^{'} = 2 x e^{3 x} + x^{2} (3 e^{3 x}) = e^{3 x} (2 x + 3 x^{2}) .$

4. Higher‑Order Derivatives

The second derivative $f^{''} (x)$ is the derivative of $f^{'} (x)$ . Higher orders are denoted $f^{(n)} (x)$ .

Position, velocity, and acceleration of Daraz delivery bike over time

Physical interpretation:

$f^{'} (x)$ = velocity (if $f$ is position).
$f^{''} (x)$ = acceleration.

Worked Example 3 – Motion of a Delivery Bike

A Daraz delivery bike’s position (km) after $t$ hours is $s (t) = 4 t^{3} - 15 t^{2} + 10 t$ .

Velocity: $v (t) = s^{'} (t) = 12 t^{2} - 30 t + 10$ .
Acceleration: $a (t) = v^{'} (t) = 24 t - 30$ .

At $t = 2$  h:

$v (2) = 12 (4) - 30 (2) + 10 = 48 - 60 + 10 = - 2$  km/h (bike is slowing).

$a (2) = 24 (2) - 30 = 48 - 30 = 18$  km/h² (positive acceleration means the bike will soon speed up again).

5. Implicit Differentiation

When a function is given implicitly, e.g., $F (x, y) = 0$ , differentiate both sides with respect to $x$ , treating $y$ as a function of $x$ (i.e., $d y / d x = y^{'}$ ).

Worked Example 4 – Circle

Given $x^{2} + y^{2} = 25$ . Find $\frac{d y}{d x}$ at the point $(3, 4)$ .

Solution

Differentiate:

$2 x + 2 y y^{'} = 0 \Rightarrow y^{'} = - \frac{x}{y}$ .

At $(3, 4)$ : $y^{'} = - \frac{3}{4}$ .

6. Applications of Differentiation

Pathao delivery cost vs. distance (minimum cost at d=0 km)

6.1 Tangent and Normal Lines

Tangent line at $x = a$ : $y = f (a) + f^{'} (a) (x - a)$ .
Normal line slope = $- 1/ f^{'} (a)$ .

6.2 Rate of Change (Real‑World)

Mobile data usage: Ncell monitors data consumption $D (t)$ (GB) over time. The derivative $D^{'} (t)$ tells the instantaneous usage rate, helping to trigger alerts when a user approaches their quota.

6.3 Optimisation (Maxima/Minima)

Find critical points where $f^{'} (x) = 0$ or undefined.
Use first‑derivative test or second‑derivative test to classify.

Worked Example 5 – Minimising Delivery Cost

A Pathao delivery charge $C (d) = 0.5 d^{2} + 20 d + 150$ (NRs) where $d$ is distance in km. Find the distance that gives minimum cost.

$C^{'} (d) = d + 20$ .
Set to zero: $d + 20 = 0 \Rightarrow d = - 20$ (not feasible).
Since the quadratic coefficient is positive, the cost is monotonically increasing for $d \geq 0$ ; the minimum occurs at the smallest possible distance, i.e., $d = 0$  km (pickup).

Interpretation: For short trips, a flat fee (150 NRs) dominates; beyond a few km the quadratic term raises cost sharply, encouraging bundling of nearby orders.

When two or more quantities change with time, differentiate the relation linking them.

Cylinder filling with water (related rates example)

Worked Example 6 – Flood Level Rise

A cylindrical water tank (radius 5 m) is being filled at $\frac{d V}{d t} = 3 m^{3} / min$ . Find the rate at which the water height $h$ rises when $h = 2$  m.

Volume $V = π r^{2} h = π (5^{2}) h = 25 π h$ .
Differentiate: $\frac{d V}{d t} = 25 π \frac{d h}{d t}$ .
Solve: $\frac{d h}{d t} = \frac{3}{25 π} \approx 0.0382 m/min$ .

7. Comparison Table – Differentiation vs. Integration

Aspect	Differentiation	Integration
Primary question	“How fast?” (instantaneous change)	“How much?” (accumulated total)
Operation	Limit of difference quotient	Antiderivative / area under curve
Linear	Yes (rules similar)	Yes
Inverse	Integration is inverse (up to constant)	Differentiation is inverse
Typical exam format	Compute $f^{'} (x)$ , find tangent, optimise, related rates	Find $\int f (x) d x$ , area, solve differential equations

8. Advantages, Disadvantages & Applications

Aspect	Advantage	Disadvantage	Typical Application
Speed of computation	Simple rules give quick results	Complex compositions may need careful chain‑rule bookkeeping	Real‑time analytics (e.g., eSewa transaction velocity)
Interpretability	Direct physical meaning (slope, rate)	May be undefined at corners/discontinuities	Engineering stress‑strain analysis
Optimization	Provides necessary conditions for extrema	Sufficiency requires second‑derivative test or global analysis	Pricing strategy in NEPSE (marginal profit)
Modeling	Handles non‑linear growth (exponential, logistic)	Closed‑form derivative may not exist for messy data	Forecasting user growth for YouTube

9. In the real world

eSewa uses the derivative of the transaction volume curve to detect spikes in payment activity, automatically scaling server resources when $d V / d t$ exceeds a threshold.
Daraz applies optimisation (first‑derivative test) to its order‑dispatch algorithm: the cost function $C (q) = 0.2 q^{2} + 15 q + 500$ (where $q$ = number of orders per batch) is differentiated to find the batch size that minimises delivery cost, improving logistics efficiency.
NEPSE analysts compute the marginal revenue $M R = d (T R) / d Q$ where total revenue $T R = P (Q) \cdot Q$ . The derivative tells investors the price change needed to increase revenue, guiding buy‑sell decisions.

Worked real example: A Nepali bank offers a loan with simple interest $I = P r t$ . The instantaneous interest rate with respect to time is $d I / d t = P r$ . If a borrower wants to know how fast interest accrues at the 3‑year mark, the bank simply evaluates $P r$ (e.g., $P = 1, 000, 000$  NRs, $r = 0.12$  yr $^{- 1}$ → $d I / d t = 120, 000$  NRs/yr). This derivative informs the borrower’s repayment schedule.

10. Common Mistakes to Avoid

Mistake	Correct Approach
Forgetting to multiply by the inner derivative in the chain rule.	Always write $f (g (x))^{'} = f^{'} (g (x)) \cdot g^{'} (x)$ .
Mixing up product vs. quotient rule signs.	Quotient: numerator derivative × denominator minus denominator derivative × numerator, all over denominator squared.
Assuming $f^{'} (x) = 0$ guarantees a maximum.	Use second‑derivative test: if $f^{''} (x) < 0$ → local maximum; if $> 0$ → local minimum.
Ignoring domain restrictions after differentiation (e.g., $ln x$ requires $x > 0$ ).	Carry forward original domain constraints.

11. Exam tip

Read the question carefully: Identify whether the problem asks for a derivative, a tangent line, a rate of change, or an optimisation condition.
Write the rule you will use before substituting numbers; this prevents algebraic slips.
For related‑rates problems, first draw a quick diagram, label all variables, write the governing equation, then differentiate with respect to time.
When the answer is a function, simplify it as much as possible; many marks are awarded for a clean final expression.
For max/min questions, always check the second derivative (or use the first‑derivative sign chart) to confirm the nature of the critical point.

In the real world

Ncell Data Alerts: Uses rate of change (derivatives) to monitor real-time data usage (D(t)) and trigger alerts when D'(t) exceeds a threshold (e.g., 90% of quota).
Pathao Delivery Optimization: Applies minimization (critical points) to bundle nearby orders (e.g., minimizing C(d) = 0.5d² + 20d + 150 NRs) to reduce delivery costs.
NTC Traffic Monitoring: Uses related rates to track vehicle speed (v(t) = s'(t)) and congestion (e.g., s(t) = 4t³ − 15t² + 10t km) on highways.

Based on the TU BITM syllabus for Basic Mathematics (MTH204), unit 4.

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Differentiation – Definitions, Rules, Applications & Real‑World Examples

Key points

1. What is a Derivative?

Visual – Tangent Slope

2. Basic Differentiation Rules

Worked Example 1 – Applying Multiple Rules

Visual – Graph of the Result

3. Differentiation of Special Functions

Worked Example 2 – Log‑Exponential Combination

4. Higher‑Order Derivatives

Worked Example 3 – Motion of a Delivery Bike

5. Implicit Differentiation

Worked Example 4 – Circle

6. Applications of Differentiation

6.1 Tangent and Normal Lines

6.2 Rate of Change (Real‑World)

6.3 Optimisation (Maxima/Minima)

Worked Example 5 – Minimising Delivery Cost

6.4 Related Rates

Worked Example 6 – Flood Level Rise

7. Comparison Table – Differentiation vs. Integration

8. Advantages, Disadvantages & Applications

9. In the real world

10. Common Mistakes to Avoid

11. Exam tip

In the real world

Discussion