Differentiation: Rules, Graphs, Business Applications

TAKEAWAYS:

• Differentiation measures instantaneous rate of change (slope of a tangent) of functions like cost, revenue, and profit.
• Basic rules (power, product, quotient, chain) simplify finding derivatives of complex functions.
• Graphical interpretation: A derivative’s sign tells whether a function is increasing/decreasing; its zero points indicate maxima/minima.
• Business applications: Differentiation optimizes profit, minimizes cost, and analyzes marginal functions (marginal cost/revenue).
• Second derivatives reveal concavity (inflection points) and acceleration in business trends.
• Exam focus: Solve problems using differentiation rules, interpret graphs, and apply to real-world scenarios (e.g., loan interest, inventory costs).

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1. Definition and Intuition

Differentiation is the process of finding the derivative of a function, which represents its instantaneous rate of change. For a function $f(x)$, the derivative $f^{\prime }(x)$ (or $\frac{df}{dx}$) is the slope of the tangent line to the curve at any point $x$.

Key Idea: From Average to Instantaneous Rate

• Average rate of change over an interval $[x_1,x_2]$: $$\text{Average rate}=\frac{f(x_2)-f(x_1)}{x_2-x_1}$$
• Instantaneous rate of change (derivative) is the limit as the interval shrinks to zero: $$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$$

Visual: The Tangent Line Approximation

Interpretation:

• The red line is the tangent to $f(x)=x^2$ at $x=1$.
• Its slope ($f^{\prime }(1)=2$) is the instantaneous rate of change at $x=1$.

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2. Basic Differentiation Rules

Master these rules to differentiate any function efficiently.

A. Power Rule

For $f(x)=x^n$, the derivative is: $$f^{\prime }(x)=n\cdot x^{n-1}$$ Example 1: Differentiate $f(x)=3x^4-5x^2+7$. $$f^{\prime }(x)=3\cdot 4x^3-5\cdot 2x^1+0=12x^3-10x$$

B. Constant Rule

The derivative of a constant $c$ is 0: $$\frac{d}{dx}[c]=0$$

C. Constant Multiple Rule

Multiply the derivative of the function by the constant: $$\frac{d}{dx}[c\cdot f(x)]=c\cdot f^{\prime }(x)$$

D. Sum/Difference Rule

Differentiate term-by-term: $$\frac{d}{dx}[f(x)\pm g(x)]=f^{\prime }(x)\pm g^{\prime }(x)$$

E. Product Rule

For two functions $u(x)$ and $v(x)$: $$\frac{d}{dx}[u\cdot v]=u^{\prime }\cdot v+u\cdot v^{\prime }$$ Example 2: Differentiate $f(x)=x^2\cdot \sin x$. Let $u=x^2$, $v=\sin x$. $$u^{\prime }=2x,v^{\prime }=\cos x$$ $$f^{\prime }(x)=2x\cdot \sin x+x^2\cdot \cos x$$

F. Quotient Rule

For $\frac{u(x)}{v(x)}$: $$\frac{d}{dx}\left[\frac{u}{v}\right]=\frac{u^{\prime }v-u{v}^{\prime }}{v^2}$$ Example 3: Differentiate $f(x)=\frac{x^2}{\ln x}$. Let $u=x^2$, $v=\ln x$. $$u^{\prime }=2x,v^{\prime }=\frac{1}{x}$$ $$f^{\prime }(x)=\frac{2x\cdot \ln x-x^2\cdot \frac{1}{x}}{(\ln x{)}^2}=\frac{2x\ln x-x}{(\ln x{)}^2}$$

G. Chain Rule (for Composite Functions)

For $f(g(x))$, the derivative is: $$f^{\prime }(g(x))\cdot g^{\prime }(x)$$ Example 4: Differentiate $f(x)=\sin (3x^2+1)$. Let $u=3x^2+1$, $f(u)=\sin u$. $$f^{\prime }(u)=\cos u,u^{\prime }=6x$$ $$f^{\prime }(x)=\cos (3x^2+1)\cdot 6x$$

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3. Graphical Interpretation of Derivatives

The derivative $f^{\prime }(x)$ tells us about the shape of $f(x)$:

Key Observations:

1. Where $f^{\prime }(x)=0$, $f(x)$ has a critical point (potential max/min).
2. The sign of $f^{\prime }(x)$ tells whether $f(x)$ is rising or falling.
3. The slope of $f^{\prime }(x)$ (i.e., $f^{\prime \prime }(x)$) tells whether $f(x)$ is concave up/down.

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4. Business Applications of Differentiation

Differentiation is used to model and optimize real-world business scenarios.

A. Marginal Functions

• Marginal Cost (MC): Derivative of the total cost function $C(x)$. $$MC=\frac{dC}{dx}$$
• Marginal Revenue (MR): Derivative of the total revenue function $R(x)$. $$MR=\frac{dR}{dx}$$
• Marginal Profit (MP): Derivative of the profit function $P(x)=R(x)-C(x)$. $$MP=\frac{dP}{dx}=MR-MC$$

Example 5: Ncell’s Data Plan Pricing Suppose Ncell’s total revenue $R(x)$ from selling $x$ GB of data is: $$R(x)=500x-0.5x^2$$ Find the marginal revenue when $x=100$ GB. $$MR=\frac{dR}{dx}=500-x$$ At $x=100$: $$MR=500-100=400\text{ (NPR per additional GB)}$$

B. Profit Maximization

Profit $P(x)=R(x)-C(x)$ is maximized where $P^{\prime }(x)=0$ (and $P^{\prime \prime }(x)<0$).

Example 6: Daraz’s Order Processing Cost Daraz’s cost function for processing $x$ orders/day is: $$C(x)=1000+20x+0.1x^2$$ Revenue from $x$ orders: $$R(x)=50x$$ Profit function: $$P(x)=R(x)-C(x)=50x-(1000+20x+0.1x^2)=-0.1x^2+30x-1000$$ Find the number of orders $x$ that maximizes profit: $$P^{\prime }(x)=-0.2x+30$$ Set $P^{\prime }(x)=0$: $$-0.2x+30=0⟹x=150\text{ orders/day}$$ Check concavity: $$P^{\prime \prime }(x)=-0.2<0\text{(Maximum)}$$ Maximum profit: $$P(150)=-0.1(150{)}^2+30(150)-1000=1625\text{ NPR}$$

C. Cost Minimization

Cost $C(x)$ is minimized where $C^{\prime }(x)=0$ (and $C^{\prime \prime }(x)>0$).

Example 7: Khalti’s Transaction Fee Khalti’s cost for processing $x$ transactions/day is: $$C(x)=500+5x+0.01x^2$$ Find the optimal number of transactions to minimize cost: $$C^{\prime }(x)=5+0.02x$$ Set $C^{\prime }(x)=0$: $$5+0.02x=0⟹x=-250\text{(Not feasible)}$$ Since $C^{\prime }(x)>0$ for all $x>0$, the cost increases with $x$. Thus, Khalti should minimize transactions to reduce costs (but this contradicts business goals, so they must balance with revenue).

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5. Second Derivative and Concavity

The second derivative $f^{\prime \prime }(x)$ tells us about the concavity of $f(x)$:

Example 8: NEPSE Stock Price Trend Suppose NEPSE’s stock price $S(t)$ over time $t$ (in months) is: $$S(t)=1000+50t-t^2$$ Find concavity and inflection points: $$S^{\prime }(t)=50-2t$$ $$S^{\prime \prime }(t)=-2<0\text{(Always concave down)}$$ No inflection points exist here.

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## In the Real World

1. eSewa’s Transaction Fees

   • Idea Used: Marginal Cost
   • How: eSewa charges a fixed fee + a small percentage per transaction. The derivative of the total cost function helps determine the optimal fee structure to maximize profit while keeping users engaged.

2. Pathao’s Driver Surge Pricing

   • Idea Used: Differentiation for Dynamic Pricing
   • How: Pathao’s algorithm differentiates demand (number of rides $x$) to adjust prices dynamically. If $D(x)=100-2x$ (demand function), the marginal revenue $MR=D^{\prime }(x)\cdot P(x)$ helps set prices during peak hours (e.g., when $x$ is high, prices surge to balance supply/demand).

3. Bank Loan Interest Calculation (NMB, Global IME)

   • Idea Used: Derivatives for Loan Optimization
   • How: Banks use differentiation to model loan repayment schedules. For a loan $L$ with interest rate $r$, the marginal cost of delay (derivative of the total repayment function) helps banks decide optimal loan terms. For example: $$\text{Total Repayment}=L(1+r{)}^t$$ The derivative with respect to time $t$ gives the instantaneous cost of delay, helping banks set penalties for late payments.

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## Exam Tip

1. Memorize Rules: Know the power, product, quotient, and chain rules by heart. Exams often test these directly.
2. Graph Interpretation: Always sketch the graph of $f(x)$ and $f^{\prime }(x)$ together. Label critical points (where $f^{\prime }(x)=0$) and state whether they are maxima/minima using the second derivative test.
3. Business Context: In word problems:
   • Identify whether you need to maximize profit ($P^{\prime }(x)=0$) or minimize cost ($C^{\prime }(x)=0$).
   • Use marginal functions (MC, MR, MP) to answer questions about optimal production levels.
4. Units Matter: Always include units in your final answer (e.g., "NPR per unit," "orders/day").
5. Check Concavity: After finding critical points, always verify with the second derivative to confirm maxima/minima.
6. Common Pitfalls:
   • Forgetting the chain rule for composite functions (e.g., $\sin (2x)$ → derivative is $2\cos (2x)$, not $\cos (2x)$).
   • Misapplying the quotient rule (remember $\frac{u^{\prime }v-u{v}^{\prime }}{v^2}$).
   • Ignoring domain restrictions (e.g., $\ln x$ is undefined for $x\le 0$).

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## Practice Problems (Solve These!)

1. Differentiate $f(x)=e^{3x}\cdot \ln x$ using the product rule.
2. Find the marginal cost for $C(x)=200+10x+0.5x^2$ when $x=50$.
3. A company’s profit function is $P(x)=-2x^3+60x^2+100x-500$. Find the values of $x$ that maximize profit.
4. Sketch the graph of $f(x)=x^3-6x^2+9x$ and identify all critical points, stating whether they are maxima or minima.
5. Real-world: Suppose Daraz’s revenue from selling $x$ units is $R(x)=100x-0.5x^2$ and cost is $C(x)=20x+100$. Find the optimal number of units to maximize profit and calculate the maximum profit.

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## Summary Table: Differentiation Rules