Application of Derivatives — Mathematics Class 12 Notes (CBSE & HBSE)
Free NCERT Mathematics notes for Application of Derivatives (Class 12) on Siksha Sarovar, aligned to CBSE and Haryana Board (HBSE). This chapter is broken into 3 topics with clear explanations, formulas, solved examples and board-pattern practice — free to read, no sign-up required.
Board exam focus — Application of Derivatives (CBSE & HBSE)
Application of Derivatives carries 8-10 marks. CBSE tests rate of change, increasing/decreasing functions, tangents and normals, and maxima/minima. The 5-mark question usually involves finding absolute maximum/minimum or a word problem on optimization.
Rate of Change and Increasing/Decreasing Functions
Rate of Change
dy/dx represents the rate of change of y with respect to x.
If y = f(x), then dy/dx|_{x=a} is the instantaneous rate of change at x = a.
Important applications:
- Velocity = ds/dt (rate of change of displacement)
- Acceleration = dv/dt = d²s/dt²
- Rate of area change: dA/dt = dA/dr · dr/dt (chain rule)
- Rate of volume change: dV/dt = dV/dr · dr/dt
Worked Example: A balloon is being inflated. If its radius increases at 2 cm/s, find the rate of increase of volume when radius = 5 cm.
- V = (4/3)πr³
- dV/dt = 4πr² · dr/dt = 4π(25)(2) = 200π cm³/s
Increasing and Decreasing Functions
| Behaviour | Condition | Implication |
|---|---|---|
| Strictly Increasing on (a,b) | f'(x) > 0 ∀ x ∈ (a,b) | f(x₁) < f(x₂) when x₁ < x₂ |
| Strictly Decreasing on (a,b) | f'(x) < 0 ∀ x ∈ (a,b) | f(x₁) > f(x₂) when x₁ < x₂ |
| Constant on (a,b) | f'(x) = 0 ∀ x ∈ (a,b) | f(x) = constant |
Method to find intervals:
- Find f'(x)
- Solve f'(x) = 0 to get critical points
- Check sign of f'(x) in each interval between critical points
CBSE Tip: Always express your answer as open intervals. Write "f is increasing on (a,b)" not "for a < x < b".
Tangents and Normals
Equations of Tangent and Normal
Slope of tangent to y = f(x) at (x₀,y₀): m = dy/dx|(x₀,y₀)
Equation of tangent: y − y₀ = m(x − x₀)
Slope of normal: −1/m (perpendicular to tangent)
Equation of normal: y − y₀ = (−1/m)(x − x₀)
Special cases:
- If m = 0 (horizontal tangent): tangent is y = y₀; normal is x = x₀
- If m → ∞ (vertical tangent): tangent is x = x₀; normal is y = y₀
Length of Tangent and Normal
| Quantity | Formula | ||
|---|---|---|---|
| Length of tangent | y₀ | √(1+(dx/dy)²) | |
| Length of normal | y₀ | √(1+(dy/dx)²) | |
| Length of subtangent | y₀/(dy/dx) | ||
| Length of subnormal | y₀·(dy/dx) |
Angle Between Two Curves
If m₁ and m₂ are slopes of tangents to two curves at their intersection: tan θ = |(m₁−m₂)/(1+m₁m₂)|
Curves are orthogonal (cut at right angles) iff m₁·m₂ = −1.
Finding Tangent Parallel to x-axis
Tangent is parallel to x-axis when dy/dx = 0 (horizontal tangent). Find x from dy/dx = 0, then find y, giving point of tangency.
CBSE Tip: For parametric curves x=f(t), y=g(t): slope = (dy/dt)/(dx/dt). Always substitute back to find the point coordinates.
Maxima and Minima
Critical Points and Extrema
Critical point: x = c where f'(c) = 0 or f'(c) doesn't exist.
First Derivative Test
- Find f'(x) = 0 ⟹ critical points c₁, c₂, …
- Check sign change of f'(x) around each critical point:
| Sign change | Nature |
|---|---|
| + to − (decreasing after c) | Local Maximum |
| − to + (increasing after c) | Local Minimum |
| No change | Point of Inflection (neither max nor min) |
Second Derivative Test
At critical point c (where f'(c) = 0):
- f''(c) < 0 ⟹ Local Maximum
- f''(c) > 0 ⟹ Local Minimum
- f''(c) = 0 ⟹ Test fails (use first derivative test)
Absolute (Global) Maxima and Minima on [a,b]
- Find all critical points in (a,b)
- Evaluate f at critical points AND at endpoints f(a), f(b)
- Largest value = absolute maximum; smallest = absolute minimum
Optimization Word Problems — Strategy
- Identify the function to be optimized
- Express in terms of ONE variable (using constraints)
- Find derivative and set to zero
- Verify it is max or min using second derivative test
- Find the actual maximum/minimum value
Example: Find two positive numbers whose sum is 15 and product is maximum.
- Let numbers be x and 15−x; Product P = x(15−x) = 15x−x²
- dP/dx = 15−2x = 0 ⟹ x = 7.5
- d²P/dx² = −2 < 0 ⟹ Maximum
- Numbers: 7.5 and 7.5, Product = 56.25
CBSE Tip: The 5-mark optimization question always has a constraint equation and an objective function. Draw a diagram, write variables, eliminate one using constraint, then differentiate.
Frequently asked questions
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Concept explanations, key formulas and definitions, fully solved examples and board-pattern practice questions for Application of Derivatives.