Chapter 6: Application of Derivatives
Complete Board Exam Focused Notes with PYQs and Problem-Solving Strategies
Exam Weightage & Blueprint
Total: ~16 MarksApplication of Derivatives is one of the most scoring chapters with direct formula-based questions. Master the working rules for guaranteed marks!
| Question Type | Marks | Frequency | Focus Topic |
|---|---|---|---|
| MCQ | 1 | Very High | Rate of change, Increasing/Decreasing intervals |
| Short Answer (2M) | 2 | High | Tangent/Normal equations, Simple maxima/minima |
| Short Answer (3M) | 3 | Very High | Increasing/Decreasing functions, Local maxima/minima |
| Long Answer (5M) | 5 | Very High | Word problems, Optimization, Absolute maxima/minima |
⏰ Last 24-Hour Checklist
Rate of Change
- Basic: $\frac{dy}{dx}$ = rate of change of $y$ w.r.t. $x$
- Chain Rule: $\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}$
- Volume/Area formulas for sphere, cone, cylinder
- Marginal cost = $\frac{dC}{dx}$, Marginal revenue = $\frac{dR}{dx}$
Increasing/Decreasing
- Increasing: $f'(x) > 0$
- Decreasing: $f'(x) < 0$
- Find critical points: $f'(x) = 0$
- Test intervals on number line
Maxima/Minima Tests
- 1st Derivative Test: Sign change of $f'(x)$
- 2nd Derivative Test: $f'(c) = 0$ and check $f''(c)$
- $f''(c) < 0$ → Local maxima
- $f''(c) > 0$ → Local minima
Absolute Max/Min
- Find all critical points in $[a,b]$
- Evaluate $f$ at critical points
- Evaluate $f$ at end points $a$ and $b$
- Compare values to find max/min
Rate of Change of Quantities ⭐⭐⭐⭐⭐
If $y = f(x)$, then $\frac{dy}{dx}\bigg|_{x=x_0}$ represents the instantaneous rate of change of $y$ at $x = x_0$.
Chain Rule for Related Rates
Important Formulas (Must Remember!) 📝
Circle
• Area: $A = \pi r^2$
• Circumference: $C = 2\pi r$
• $\frac{dA}{dr} = 2\pi r$
Sphere
• Volume: $V = \frac{4}{3}\pi r^3$
• Surface Area: $S = 4\pi r^2$
• $\frac{dV}{dr} = 4\pi r^2$
Cube
• Volume: $V = x^3$
• Surface Area: $S = 6x^2$
• $\frac{dV}{dx} = 3x^2$
Cone
• Volume: $V = \frac{1}{3}\pi r^2 h$
• Curved Surface: $S = \pi r l$
• $l = \sqrt{r^2 + h^2}$
Common Applications
• Marginal Cost: $MC = \frac{dC}{dx}$ (rate of change of total cost)
• Marginal Revenue: $MR = \frac{dR}{dx}$ (rate of change of total revenue)
• Marginal Profit: $MP = MR - MC$
Increasing & Decreasing Functions 🔥🔥🔥
- $f$ is increasing on $I$ if $x_1 < x_2$ in $I \Rightarrow f(x_1) \leq f(x_2)$
- $f$ is decreasing on $I$ if $x_1 < x_2$ in $I \Rightarrow f(x_1) \geq f(x_2)$
- $f$ is strictly increasing if $x_1 < x_2 \Rightarrow f(x_1) < f(x_2)$
- $f$ is strictly decreasing if $x_1 < x_2 \Rightarrow f(x_1) > f(x_2)$
First Derivative Test for Monotonicity
• $f$ is increasing on $[a,b]$ if $f'(x) \geq 0$ for all $x \in (a,b)$
• $f$ is decreasing on $[a,b]$ if $f'(x) \leq 0$ for all $x \in (a,b)$
• $f$ is constant on $[a,b]$ if $f'(x) = 0$ for all $x \in (a,b)$
Step-by-Step Working Rule
Step 1: Find $f'(x)$
Step 2: Solve $f'(x) = 0$ to find critical points
Step 3: Plot critical points on number line to divide into intervals
Step 4: Check sign of $f'(x)$ in each interval (pick test points)
Step 5: If $f'(x) > 0$ → increasing; if $f'(x) < 0$ → decreasing
Example: If $f'(x) = (x-1)(x+2)(x-3)$, check signs at $x = -3, 0, 2, 4$:
• At $x = -3$: $(-)(-)(-) = - $ → Decreasing
• At $x = 0$: $(-)(+)(-) = +$ → Increasing
• At $x = 2$: $(+)(+)(-) = -$ → Decreasing
• At $x = 4$: $(+)(+)(+) = +$ → Increasing
Maxima and Minima 🎯🎯🎯
Key Definitions
Local Minima: $c$ is a point of local minima if there exists $h > 0$ such that: $$f(c) \leq f(x) \text{ for all } x \in (c-h, c+h)$$
Critical Point: A point $c$ where either $f'(c) = 0$ or $f$ is not differentiable.
First Derivative Test
| Sign of $f'(x)$ | Left of $c$ | Right of $c$ | Nature at $c$ |
|---|---|---|---|
| Case 1 | + | − | Local Maxima |
| Case 2 | − | + | Local Minima |
| Case 3 | + or − | Same sign | Neither (Point of Inflection) |
Second Derivative Test
• If $f''(c) < 0$ → $c$ is point of local maxima
• If $f''(c) > 0$ → $c$ is point of local minima
• If $f''(c) = 0$ → Test fails, use First Derivative Test
• $f''(c) < 0$ → Graph is ∩ (concave down) → Maximum at top
• $f''(c) > 0$ → Graph is ∪ (concave up) → Minimum at bottom
Working Rule for Local Maxima/Minima
Step 1: Find $f'(x)$ and solve $f'(x) = 0$ for critical points
Step 2: For each critical point $c$, use either:
• Method 1 (2nd Derivative Test): Find $f''(c)$ and check sign
• Method 2 (1st Derivative Test): Check sign of $f'(x)$ left/right of $c$
Step 3: Calculate $f(c)$ to find local maximum/minimum value
Absolute Maximum and Minimum Values
• Absolute Maximum: Greatest value of $f$ on entire interval $[a,b]$
• Absolute Minimum: Least value of $f$ on entire interval $[a,b]$
Working Rule (Most Important!) 🔥
Step 1: Find all critical points in $(a,b)$ where $f'(x) = 0$ or $f'(x)$ doesn't exist
Step 2: List all critical points: $c_1, c_2, c_3, ...$
Step 3: Calculate: $f(a), f(c_1), f(c_2), ..., f(b)$
Step 4: The largest value is absolute maximum
Step 5: The smallest value is absolute minimum
• Local max/min: Highest/lowest in a small neighborhood
• Absolute max/min: Highest/lowest in the entire interval
• Absolute max/min can occur at endpoints or at critical points!
Important Theorems
Theorem 2: If $f$ is differentiable and attains absolute max/min at interior point $c$, then $f'(c) = 0$.
Solved Examples (Board Pattern)
Q1. The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference? (2 Marks)
Let $r$ be the radius and $C$ be the circumference. Then $C = 2\pi r$
Differentiating w.r.t. time $t$:
$\frac{dC}{dt} = 2\pi \frac{dr}{dt}$
Given: $\frac{dr}{dt} = 0.7$ cm/s
Therefore: $\frac{dC}{dt} = 2\pi(0.7) = 1.4\pi$ cm/s
Answer: The circumference is increasing at $1.4\pi$ cm/s or approximately $4.4$ cm/s.
Q2. Find the intervals in which $f(x) = 2x^3 - 3x^2 - 36x + 7$ is (a) increasing (b) decreasing. (3 Marks)
$f'(x) = 6x^2 - 6x - 36 = 6(x^2 - x - 6) = 6(x-3)(x+2)$
$f'(x) = 0 \Rightarrow x = 3$ or $x = -2$
These points divide real line into intervals: $(-\infty, -2), (-2, 3), (3, \infty)$
• In $(-\infty, -2)$: Take $x = -3$, $f'(-3) = 6(-6)(-5) = 180 > 0$ → Increasing
• In $(-2, 3)$: Take $x = 0$, $f'(0) = 6(-3)(2) = -36 < 0$ → Decreasing
• In $(3, \infty)$: Take $x = 4$, $f'(4) = 6(1)(6) = 36 > 0$ → Increasing
(a) $f$ is increasing on $(-\infty, -2) \cup (3, \infty)$
(b) $f$ is decreasing on $(-2, 3)$
Q3. Find the local maxima and local minima of $f(x) = x^3 - 3x^2 + 4x$. (3 Marks)
$f'(x) = 3x^2 - 6x + 4$
For critical points: $f'(x) = 0$
$3x^2 - 6x + 4 = 0$
Discriminant: $D = 36 - 48 = -12 < 0$
No real roots, so no critical points.
$f'(x) = 3(x^2 - 2x + \frac{4}{3}) = 3[(x-1)^2 + \frac{1}{3}] > 0$ for all $x \in \mathbb{R}$
Since $f'(x) > 0$ for all $x$, the function is strictly increasing on $\mathbb{R}$.
Therefore: No local maxima or minima exist.
Q4. Find the absolute maximum and minimum values of $f(x) = x^3$ on $[-2, 2]$. (3 Marks)
$f'(x) = 3x^2$
$f'(x) = 0 \Rightarrow x = 0$
Critical point: $x = 0$ (lies in $[-2, 2]$)
$f(-2) = (-2)^3 = -8$
$f(0) = 0^3 = 0$
$f(2) = 2^3 = 8$
Absolute Maximum: $8$ at $x = 2$
Absolute Minimum: $-8$ at $x = -2$
Word Problems & Applications 🎯
Type 1: Optimization Problems (Most Important!)
1. Read problem carefully and identify what to maximize/minimize
2. Draw diagram if needed
3. Express the quantity to be optimized as function of one variable
4. Find derivative and critical points
5. Use first or second derivative test
6. Answer in context with proper units
Classic Problem Types
Type A: Two Numbers
Problem: Find two positive numbers whose sum is $k$ and product is maximum.
Answer: Both numbers = $\frac{k}{2}$
Method: Let numbers be $x$ and $k-x$. Maximize $P = x(k-x)$
Type B: Rectangle in Circle
Problem: Rectangle of maximum area inscribed in circle of radius $r$.
Answer: Square with side $r\sqrt{2}$
Max Area: $2r^2$
Type C: Open Box
Problem: Box from rectangular sheet by cutting squares from corners.
Method: If cut $x$ cm, volume $V = x(l-2x)(b-2x)$
Find $\frac{dV}{dx} = 0$ and solve
Type D: Cylinder in Sphere
Problem: Right circular cylinder of max volume in sphere of radius $R$.
Answer: Height = $\frac{2R}{\sqrt{3}}$
Max Volume: $\frac{4\pi R^3}{3\sqrt{3}}$
Solved Application Problem
Q. Find two positive numbers whose sum is 16 and sum of whose cubes is minimum. (5 Marks)
Solution:
Step 1: Let the two numbers be $x$ and $16-x$ where $0 < x < 16$
Step 2: Sum of cubes: $S = x^3 + (16-x)^3$
Step 3: Differentiate: $\frac{dS}{dx} = 3x^2 - 3(16-x)^2 = 3[x^2 - (16-x)^2]$ $= 3[x^2 - 256 + 32x - x^2] = 3(32x - 256) = 96(x - 8)$
Step 4: For critical points: $\frac{dS}{dx} = 0 \Rightarrow x = 8$
Step 5: Second derivative: $\frac{d^2S}{dx^2} = 96 > 0$ → Minimum at $x = 8$
Answer: The two numbers are $8$ and $8$.
Previous Year Questions (PYQs)
(A) increasing on $\mathbb{R}$
(B) decreasing on $\mathbb{R}$
(C) increasing on $(-\infty, -1) \cup (1, \infty)$
(D) decreasing on $(-1, 1)$
Ans: (C) and (D) both. $f'(x) = 3x^2 - 3 = 3(x-1)(x+1)$.
Sign analysis shows increasing on $(-\infty, -1) \cup (1, \infty)$ and decreasing on $(-1, 1)$.
Solution:
Volume $V = x^3$, Surface area $S = 6x^2$
Given: $\frac{dV}{dt} = 8$ cm³/s
$\frac{dV}{dt} = 3x^2\frac{dx}{dt} \Rightarrow 8 = 3(12)^2\frac{dx}{dt}$
$\frac{dx}{dt} = \frac{8}{432} = \frac{1}{54}$ cm/s
$\frac{dS}{dt} = 12x\frac{dx}{dt} = 12(12)\left(\frac{1}{54}\right) = \frac{144}{54} = \frac{8}{3}$ cm²/s
Answer: $\frac{8}{3}$ cm²/s or $2.67$ cm²/s
Solution:
$f(x) = \log(\sin x)$
$f'(x) = \frac{1}{\sin x} \cdot \cos x = \cot x$
In $(0, \frac{\pi}{2})$: $\cot x > 0$, so $f'(x) > 0$ → increasing
In $(\frac{\pi}{2}, \pi)$: $\cot x < 0$, so $f'(x) < 0$ → decreasing
Hence proved.
Solution Outline:
1. Let width = $2r$, height of rectangle = $h$
2. Perimeter: $2h + 2r + \pi r = 10$
3. Area (light): $A = 2rh + \frac{\pi r^2}{2}$
4. Express $h$ from perimeter: $h = \frac{10 - 2r - \pi r}{2}$
5. Substitute in $A$ and maximize
6. Find $\frac{dA}{dr} = 0$ and solve
Answer: Width $= \frac{20}{4+\pi}$ m, Height $= \frac{10}{4+\pi}$ m
Key Steps:
1. If cylinder has radius $r$ and height $h$, then from sphere: $r^2 + \frac{h^2}{4} = R^2$
2. Volume: $V = \pi r^2 h = \pi(R^2 - \frac{h^2}{4})h = \pi R^2h - \frac{\pi h^3}{4}$
3. $\frac{dV}{dh} = \pi R^2 - \frac{3\pi h^2}{4} = 0$
4. Solving: $h^2 = \frac{4R^2}{3} \Rightarrow h = \frac{2R}{\sqrt{3}}$
5. Verify using second derivative test: $\frac{d^2V}{dh^2} < 0$ → Maximum
Common Mistakes & Scoring Tips
Common Mistakes 🚨
Scoring Tips 🏆
Formula Sheet (Quick Revision) 📋
Rate of Change
1. Rate of change of $y$ w.r.t. $x$: $\frac{dy}{dx}$2. Chain Rule: $\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}$ or $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$
3. Marginal Cost: $MC = \frac{dC}{dx}$
4. Marginal Revenue: $MR = \frac{dR}{dx}$
Increasing/Decreasing
5. $f$ increasing on $(a,b)$ if $f'(x) > 0$ for all $x \in (a,b)$6. $f$ decreasing on $(a,b)$ if $f'(x) < 0$ for all $x \in (a,b)$
7. Critical points: Solve $f'(x) = 0$
Maxima/Minima
8. First Derivative Test:• $f'(x)$ changes from + to − at $c$ → Local maximum at $c$
• $f'(x)$ changes from − to + at $c$ → Local minimum at $c$
9. Second Derivative Test:
• If $f'(c) = 0$ and $f''(c) < 0$ → Local maximum at $c$
• If $f'(c) = 0$ and $f''(c) > 0$ → Local minimum at $c$
• If $f'(c) = 0$ and $f''(c) = 0$ → Test fails, use first derivative test
Absolute Max/Min on [a,b]
10. Find all critical points in $(a,b)$11. Evaluate $f$ at critical points and at $a, b$
12. Largest value = Absolute maximum
13. Smallest value = Absolute minimum
Geometry Formulas
14. Circle: $A = \pi r^2$, $C = 2\pi r$15. Sphere: $V = \frac{4}{3}\pi r^3$, $S = 4\pi r^2$
16. Cone: $V = \frac{1}{3}\pi r^2 h$, $S = \pi rl$, $l = \sqrt{r^2 + h^2}$
17. Cylinder: $V = \pi r^2 h$, $S = 2\pi rh + 2\pi r^2$
18. Cube: $V = x^3$, $S = 6x^2$
Practice Problems (Self-Test)
Level 1: Basic (2 Marks Each)
Q1. The side of a square is increasing at the rate of 0.2 cm/s. Find the rate of increase of its perimeter.
Q2. Show that $f(x) = e^{2x}$ is increasing on $\mathbb{R}$.
Q3. Find the slope of the tangent to the curve $y = x^3 - x$ at $x = 2$.
Level 2: Intermediate (3 Marks Each)
Q4. Find the intervals in which $f(x) = x^4 - 4x^3 + 4x^2 + 15$ is increasing or decreasing.
Q5. Find all points of local maxima and minima of $f(x) = x^3 - 6x^2 + 9x + 15$.
Q6. Find the absolute maximum and minimum values of $f(x) = \sin x + \cos x$ on $[0, \pi]$.
Level 3: Advanced (5 Marks Each)
Q7. A wire of length 28 m is to be cut into two pieces. One piece is bent into a square and the other into a circle. What should be the lengths so that the combined area is minimum?
Hint: If square side is $x$ and circle radius is $r$, then $4x + 2\pi r = 28$.
Q8. Show that of all rectangles inscribed in a given fixed circle, the square has maximum area.
Q9. A tank with rectangular base and rectangular sides, open at the top, is to be constructed so that its depth is 2 m and volume is 8 m³. If building costs Rs 70/m² for base and Rs 45/m² for sides, find the cost of the least expensive tank.
Exam Strategy & Time Management
Topic-wise Time Allocation (for 80 marks paper)
| Topic | Expected Questions | Time to Allocate | Difficulty |
|---|---|---|---|
| Rate of Change | 1 MCQ + 1 Short (2M) | 5-7 minutes | Easy 🟢 |
| Increasing/Decreasing | 1 Short (3M) | 8-10 minutes | Medium 🟡 |
| Local Maxima/Minima | 1 Short (3M) | 8-10 minutes | Medium 🟡 |
| Absolute Max/Min or Application | 1 Long (5M) | 15-18 minutes | Hard 🔴 |
Last Week Revision Strategy
Day 7-5 Before Exam
✓ Revise all formulas and theorems
✓ Solve 5 PYQs from each topic
✓ Practice rate of change word problems
✓ Master the working rules
Day 4-2 Before Exam
✓ Focus on optimization problems
✓ Solve full-length problems (5M each)
✓ Revise common mistakes list
✓ Practice mental calculation of derivatives
Last Day Before Exam
✓ Go through formula sheet 3 times
✓ Solve 2 MCQs, 2 Short answers, 1 Long answer
✓ Revise the 24-hour checklist
✓ Sleep early - fresh mind = fewer calculation errors!
1. Define variables clearly (0.5 marks)
2. Form the main equation (1 mark)
3. Differentiate correctly (1 mark)
This gives you 2.5/5 marks minimum!
Important Results & Quick Facts
- If $f$ is continuous on $[a,b]$, it has both absolute max and min
- Every monotonic function attains extreme values only at endpoints
- If $f'(x) > 0$ everywhere except at isolated points where $f'(x) = 0$, then $f$ is still increasing
- Point of inflection: $f'(c) = 0$ but no extremum (e.g., $x^3$ at $x=0$)
Standard Results for Quick Solving
| Problem Type | Standard Result |
|---|---|
| Two numbers with sum $s$ and max product | Both equal to $\frac{s}{2}$ |
| Two numbers with product $p$ and min sum | Both equal to $\sqrt{p}$ |
| Rectangle of max area in circle of radius $r$ | Square with side $r\sqrt{2}$, Area $= 2r^2$ |
| Cylinder of max volume in sphere of radius $R$ | Height $= \frac{2R}{\sqrt{3}}$ |
| Cone of max volume in sphere of radius $R$ | Height $= \frac{4R}{3}$, Volume $= \frac{8\pi R^3}{27}$ |
| Cylinder of given surface and max volume | Height = Diameter of base |
Sign Testing Quick Reference
For $f'(x) = (x-a)(x-b)(x-c)$ where $a < b < c$:
| Interval | Sign of each factor | Sign of $f'(x)$ | Nature |
|---|---|---|---|
| $(-\infty, a)$ | $(-)(-)(-)$ | $-$ | Decreasing |
| $(a, b)$ | $(+)(-)(-)$ | $+$ | Increasing |
| $(b, c)$ | $(+)(+)(-)$ | $-$ | Decreasing |
| $(c, \infty)$ | $(+)(+)(+)$ | $+$ | Increasing |