1. Applications of Derivatives
1.1 Rate of Change
Derivative dy/dx = rate of change of y with respect to x. Business mein cost, revenue, profit ke changes calculate karte hain.
Marginal Cost (MC) = dC/dq (cost ka derivative w.r.t. quantity)
Marginal Revenue (MR) = dR/dq
Marginal Profit (MP) = dP/dq = MR − MC
Udaharan: Marginal Cost
Total Cost C(q) = q³ − 6q² + 15q + 10
MC = dC/dq = 3q² − 12q + 15
At q=3: MC = 3(9) − 12(3) + 15 = 27 − 36 + 15 = 6
Matlab: 3rd unit produce karne ka marginal cost ₹6 hai.
1.2 Increasing & Decreasing Functions
f'(x) > 0 for x ∈ (a,b) → f increasing on (a,b)
f'(x) < 0 for x ∈ (a,b) → f decreasing on (a,b)
f'(x) = 0 at x = c → c is a critical point
Udaharan
f(x) = x² − 4x + 3
f'(x) = 2x − 4
f'(x) = 0 → x = 2
x < 2: f'(x) < 0 → decreasing
x > 2: f'(x) > 0 → increasing
1.3 Maxima & Minima
First Derivative Test:
f'(c) = 0 aur f'(x) +ve se -ve ho → Local Maximum at x=c
f'(c) = 0 aur f'(x) -ve se +ve ho → Local Minimum at x=c
Second Derivative Test:
f'(c) = 0 aur f''(c) < 0 → Local Maximum
f'(c) = 0 aur f''(c) > 0 → Local Minimum
f'(c) = 0 aur f''(c) = 0 → Inconclusive
Udaharan: Profit Maximization
Profit P(q) = −q² + 10q − 16
P'(q) = −2q + 10 = 0 → q = 5
P''(q) = −2 < 0 → Maximum at q=5
Max Profit = −25 + 50 − 16 = ₹9
Isliye 5 units produce karne se maximum profit hoga!
1.4 Elasticity of Demand
Price Elasticity of Demand (PED) = (dq/dp) × (p/q)
|PED| > 1 → Elastic demand (price change → bada quantity change)
|PED| < 1 → Inelastic demand
|PED| = 1 → Unit elastic
Udaharan: Elasticity
Demand: q = 100 − 5p
dq/dp = −5
At p=10: q = 100−50 = 50
PED = −5 × (10/50) = −1
|PED| = 1 → Unit elastic demand
2. Integration
2.1 Basic Integration Formulas
| Function f(x) | ∫f(x)dx | Condition |
|---|
| xⁿ | xⁿ⁺¹/(n+1) + C | n ≠ −1 |
| 1/x | ln|x| + C | x ≠ 0 |
| eˣ | eˣ + C | — |
| aˣ | aˣ/ln(a) + C | a>0, a≠1 |
| sin x | −cos x + C | — |
| cos x | sin x + C | — |
| 1/(1+x²) | tan⁻¹x + C | — |
2.2 Definite Integral — Area Under Curve
∫[a to b] f(x) dx = F(b) − F(a) where F'(x) = f(x)
Area = |∫[a to b] f(x) dx| (negative area bhi positive count hoti hai)
Udaharan: Area Calculation
Area under f(x) = x² from x=0 to x=3:
∫₀³ x² dx = [x³/3]₀³ = 27/3 − 0 = 9 sq units
2.3 Consumer & Producer Surplus
Consumer Surplus = ∫[0 to q₀] D(q) dq − p₀×q₀
Producer Surplus = p₀×q₀ − ∫[0 to q₀] S(q) dq
Where D(q) = demand function, S(q) = supply function
p₀,q₀ = equilibrium price & quantity
Udaharan: Consumer Surplus
Demand: p = 12 − 2q, Equilibrium: q₀=4, p₀=4
CS = ∫₀⁴(12−2q)dq − 4×4
= [12q − q²]₀⁴ − 16
= (48−16) − 16 = 32 − 16 = 16
3. Differential Equations
3.1 Basic Concepts
Order: Highest derivative ki power (dy/dx = order 1, d²y/dx² = order 2)
Degree: Highest derivative ka power (after clearing fractions/radicals)
General solution: Arbitrary constants ke saath
Particular solution: Initial conditions use karke constants find karo
3.2 Variable Separable Method
dy/dx = f(x)·g(y) type equations:
Step 1: Separate: dy/g(y) = f(x)dx
Step 2: Integrate both sides: ∫dy/g(y) = ∫f(x)dx + C
Udaharan: Growth Model
dN/dt = kN (population growth model)
dN/N = k dt
∫dN/N = ∫k dt
ln N = kt + C
N = N₀eᵏᵗ (exponential growth!)
If N₀=1000, k=0.05: N at t=10 = 1000×e^0.5 ≈ 1649
3.3 Homogeneous Differential Equations
dy/dx = f(y/x) type → Substitution: y = vx → dy/dx = v + x(dv/dx)
Converts to variable separable form!
3.4 Applications
| Model | Equation | Solution |
|---|
| Exponential Growth | dy/dt = ky | y = y₀eᵏᵗ |
| Exponential Decay | dy/dt = −ky | y = y₀e⁻ᵏᵗ |
| Newton's Cooling | dT/dt = −k(T−Tₐ) | T = Tₐ+(T₀−Tₐ)e⁻ᵏᵗ |
| Logistic Growth | dy/dt = ky(1−y/K) | S-curve (sigmoid) |
📝 Formula Summary
- MC = dC/dq, MR = dR/dq, MP = MR − MC
- Local max: f'=0, f''<0; Local min: f'=0, f''>0
- PED = (dq/dp)·(p/q)
- ∫xⁿdx = xⁿ⁺¹/(n+1)+C, ∫(1/x)dx = ln|x|+C
- Exponential: dy/dt = ky → y = y₀eᵏᵗ