Calculus I – Differentiation, Rules, Marginals & Applications

1. Limits and Continuity – the foundation of differentiation

A limit describes the value that a function $f(x)$ approaches as the argument $x$ approaches a point $a$.
$$\underset{x\to a}{\lim }f(x)=L$$
If the limit exists and equals the function value $f(a)$, the function is continuous at $a$. Continuity is a prerequisite for differentiability.

Visual: Limit of a linear demand curve

The dotted point marks $\underset{p\to 50}{\lim }(200-2p)=100$.

2. Definition of the Derivative

The derivative of $f$ at $x=a$ is the limit of the average rate of change: $$f^{\prime }(a)=\underset{h\to 0}{\lim }\frac{f(a+h)-f(a)}{h}$$
Geometrically, $f^{\prime }(a)$ is the slope of the tangent line to the curve at $(a,f(a))$.

Visual: Tangent line to a profit curve

The red line is the tangent; its slope ${\pi }^{\prime }(20)=-20$ represents marginal profit.

3. Basic Differentiation Rules

Visual: Power rule graph

4. Implicit Differentiation

When a relation is given implicitly, e.g. $x^2+y^2=R^2$ (circle), differentiate both sides with respect to $x$ treating $y$ as a function of $x$: $$2x+2y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac{x}{y}$$
In business, implicit differentiation helps when cost and output are linked by a non‑explicit equation.

5. Partial Differentiation – multivariate production functions

A partial derivative measures the rate of change of a multivariable function with respect to one variable while holding the others constant.

For the Cobb‑Douglas production function
$$Q(L,K)=A{L}^{\alpha }{K}^{\beta }$$
the partial derivatives are
$$\frac{\partial Q}{\partial L}=A\alpha L^{\alpha -1}{K}^{\beta },\frac{\partial Q}{\partial K}=A\beta L^{\alpha }{K}^{\beta -1}$$

Worked Example (exam style)

Problem: A firm’s production function is $Q=100L^{0.5}{K}^{0.5}$. Capital $K$ is increased by 5 % and labour $L$ is decreased by 2 %. Estimate the percentage change in output using partial differentiation.

Solution:

1. Compute partial derivatives at the current point $(L_0,K_0)$: $$\frac{\partial Q}{\partial L}=100\cdot 0.5L^{-0.5}{K}^{0.5}=50\frac{K^{0.5}}{L^{0.5}}$$ $$\frac{\partial Q}{\partial K}=50\frac{L^{0.5}}{K^{0.5}}$$

2. Express differential change: $$dQ=\frac{\partial Q}{\partial L}dL+\frac{\partial Q}{\partial K}dK$$

3. Relative (percentage) change: $$\frac{dQ}{Q}=\frac{1}{2}\frac{dL}{L}+\frac{1}{2}\frac{dK}{K}$$ (Because for Cobb‑Douglas with equal exponents the elasticity of each input equals its exponent.)

4. Insert the given changes: $\frac{dL}{L}=-0.02$, $\frac{dK}{K}=+0.05$: $$\frac{dQ}{Q}=\frac{1}{2}(-0.02)+\frac{1}{2}(0.05)=-0.01+0.025=0.015$$

5. Result: Output rises by approximately 1.5 %.

6. Business Applications of the Derivative

6.1 Marginal Analysis

• Marginal Cost (MC): $MC(q)=\frac{dC}{dq}$ – cost of producing one additional unit.
• Marginal Revenue (MR): $MR(q)=\frac{dR}{dq}$ – revenue from one additional unit.
• Marginal Profit (MP): $MP(q)=MR(q)-MC(q)$.

Example: Linear cost and quadratic revenue

Cost: $C(q)=20+4q$
Revenue: $R(q)=10q-0.1q^2$

$$MC=4,MR=10-0.2q,MP=(10-0.2q)-4=6-0.2q$$

Set $MP=0$ → $q^{\ast }=30$ units maximises profit.

Visual: MC, MR, MP curves

The intersection of MR and MC (green meets blue) occurs at $q=30$.

6.2 Optimization

To find the optimum, compute first derivative, set to zero (critical point), then use second derivative test:

flowchart TD
    A["Total Cost: C(q) = 0.05q² - 3q + 200"] --> B["First Derivative: C'(q) = 0.1q - 3"]
    B --> C["Set C'(q) = 0 → q = 30"]
    C --> D["Second Derivative: C''(q) = 0.1 > 0 → Minimum"]
    D --> E["Optimal Quantity: q = 30 units"]

• If $f^{\prime \prime }(x)>0$ → local minimum.
• If $f^{\prime \prime }(x)<0$ → local maximum.

Example: Minimising total cost with a discount

Total cost with bulk discount: $C(q)=0.05q^2-3q+200$.

$C^{\prime }(q)=0.10q-3$. Set to zero → $q=30$.

$C^{\prime \prime }(q)=0.10>0$ → minimum cost at 30 units.

6.3 Related‑Rates

When two quantities change with time, differentiate the relationship with respect to $t$.

Example: Delivery speed and distance

A courier travels along a straight road. Distance $s(t)=vt$ where speed $v$ is increasing at $0.2{\text{km/h}}^2$. If at $t=2$ h the speed is $40$ km/h, the rate of change of distance is: $$\frac{ds}{dt}=v=40\text{km/h}$$ and the acceleration $\frac{dv}{dt}=0.2{\text{km/h}}^2$.

7. Comparison Table – Differentiation vs. Integration in Business

| Aspect                | Differentiation (Rate)                     | Integration (Accumulation)                     |
|-----------------------|--------------------------------------------|-----------------------------------------------|
| Primary question      | “How fast is something changing?”          | “What is the total over a period?”            |
| Typical output        | Marginal cost, marginal revenue            | Total cost, total revenue over a time span    |
| Key formula           | \(f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\) | \(\displaystyle\int_{a}^{b} f(x)\,dx\)        |
| Business use‑case     | Pricing decisions, break‑even analysis    | Cumulative cash‑flow, depreciation schedules  |
| Computational tool   | Derivative rules, partial derivatives      | Antiderivatives, definite integrals           |

8. Advantages & Limitations of Using Calculus in Business

9. In the real world

1. eSewa transaction fee – eSewa charges a flat 2 % fee plus a fixed Rs 10 per transaction. The marginal fee with respect to transaction amount $A$ is the derivative of the fee function $F(A)=0.02A+10$; thus $F^{\prime }(A)=0.02$. This constant marginal fee tells merchants that each extra rupee transferred adds exactly 2 paise to the cost, guiding pricing decisions for online sales.

2. Daraz order‑processing queue – Daraz models the average waiting time $W(q)=\frac{q}{\mu -\lambda }$ where $q$ is the number of orders, $\lambda$ the arrival rate and $\mu$ the service rate. Differentiating $W$ with respect to $q$ gives $\frac{dW}{dq}=\frac{1}{\mu -\lambda }$, the marginal increase in waiting time per additional order. Management uses this to decide when to add extra fulfillment staff (increase $\mu$).

3. Ncell data‑usage billing – Ncell offers a plan where the first 5 GB cost Rs 500, and each additional GB costs Rs 50. The cost function $C(x)=500+50\max (0,x-5)$ has a derivative $C^{\prime }(x)=0$ for $x\le 5$ and $C^{\prime }(x)=50$ for $x>5$. The jump in marginal cost at 5 GB explains why many customers limit usage just below the threshold, a behaviour that Ncell exploits in promotional bundles.

Worked tie‑in: The partial‑derivative example above (production function) mirrors a manufacturing plant that can increase capital equipment by 5 % (new machines) while reducing labour hours by 2 % (automation). The 1.5 % rise in output predicts the plant’s decision to invest in machinery despite a slight workforce cut.

10. Summary of Key Steps for Solving Differentiation Problems

1. Identify the function type (explicit, implicit, multivariate).
2. Choose the appropriate rule (power, product, chain, etc.).
3. Apply the rule step‑by‑step, simplifying algebraically.
4. For marginal analysis, interpret the derivative in business terms.
5. For optimisation, set first derivative to zero, test second derivative.
6. For related rates, differentiate with respect to time, substitute known rates.

In the real world

• eSewa: Uses related-rates to dynamically adjust transaction fees based on the rate of change of transaction volume. If transactions spike during festivals, the system calculates the marginal increase in processing costs and adjusts fees accordingly.
• NTC (Nepal Telecom): Applies marginal cost analysis to determine the optimal number of new base stations to install in a region. The derivative of the cost function (C(q)) helps decide where the marginal cost of adding a station equals the marginal benefit of improved coverage.
• Nepal Rastra Bank: Employs partial differentiation in the Cobb-Douglas production function to analyze how changes in labor (L) and capital (K) inputs affect GDP growth. For example, if labor increases by 3% and capital by 2%, the bank estimates the combined effect on output using partial derivatives.

Exam tip

• Mark the method first: In any question, write a short line “Use product rule” or “Apply partial differentiation” before the algebra; examiners award points for the correct approach even if arithmetic slips.
• Keep units visible: Write “Rs per unit” for marginal cost, “units per hour” for related‑rate speed, etc.; it prevents careless sign errors.
• Check the second‑derivative sign for optimisation questions – a common source of lost marks.
• For Cobb‑Douglas functions, remember that the exponent equals the output elasticity; you can shortcut the partial‑derivative calculation by using $\frac{\partial Q}{\partial L}=\alpha \frac{Q}{L}$.