NCERT Plus Two (Class 12) Mathematics • Chapter 6

Application of Derivatives

Q: Find the rate of change of the area of a circle with respect to its radius r when r = 3 cm .

A = πr² ⇒ dA dr = 2πr . At r = 3 cm : dA dr = 2π(3) = 6π cm²/cm .

Q: The total revenue from selling x units of a product is R(x) = 13x² + 26x + 15 . Find the marginal revenue when x = 7 .

Marginal Revenue = R'(x) = 26x + 26 . At x = 7 : R'(7) = 26(7) + 26 = 182 + 26 = ₹ 208 .

Q: Find the intervals in which f(x) = 2x³ − 3x² − 36x + 7 is (a) strictly increasing (b) strictly decreasing.

f'(x) = 6x² − 6x − 36 = 6(x − 3)(x + 2) ; critical points x = −2, 3 . • Strictly Increasing: (−∞, −2) and (3, ∞) (where f'(x) > 0 ). • Strictly Decreasing: (−2, 3) (where f'(x) < 0 ).

Q: Prove that f(x) = x − sin x is an increasing function on ℝ .

f'(x) = 1 − cos x . Since −1 ≤ cos x ≤ 1 for all x ∈ ℝ , we have 1 − cos x ≥ 0 everywhere. Hence f(x) is increasing on ℝ .

Q: Prove that y = 4 sin θ 2 + cos θ − θ is an increasing function of θ in [0, π/2] .

dy dθ = (2+cos θ)(4 cos θ) − 4 sin θ(−sin θ) (2 + cos θ)² − 1 = 8 cos θ + 4 − (4 + 4 cos θ + cos²θ) (2 + cos θ)² = cos θ (4 − cos θ) (2 + cos θ)² . In [0, π/2] , cos θ ≥ 0 and 4 − cos θ > 0 , so dy/dθ ≥ 0 . Verified!

Q: Find all points of local maxima and local minima of f(x) = x³ − 6x² + 9x + 15 .

f'(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3) = 0 ⇒ x = 1, 3 . f''(x) = 6x − 12 . • At x = 1 : f''(1) = −6 0 ⇒ Local Min = 15 .

Q: Find the absolute maximum and absolute minimum values of f(x) = sin x + cos x , x ∈ [0, π] .

f'(x) = cos x − sin x = 0 ⇒ tan x = 1 ⇒ x = π/4 ∈ [0, π] . Evaluate candidates: f(0) = 1 , f(π/4) = 1 √2 + 1 √2 = √2 , f(π) = −1 . • Absolute Max = √2 (at π/4 ) | Absolute Min = −1 (at π ).

Q: A square piece of tin of side 18 cm is made into an open box by cutting equal squares from each corner and folding up flaps. What side square should be cut so the box volume is maximum?

Let cut square side be x ( 0 < x < 9 ). V(x) = x(18 − 2x)² = 4(x³ − 18x² + 81x) . V'(x) = 4(3x² − 36x + 81) = 12(x − 3)(x − 9) = 0 ⇒ x = 3 cm ( x = 9 impossible). V''(x) = 12(2x − 12) ⇒ V''(3) = −72 < 0 (Max!). Cut side = 3 cm ; Max Volume = 3 × 12² = 432 cm³ .

Related rates of change, increasing & decreasing intervals, critical points, first & second derivative tests, absolute extrema, and real-life optimisation.

📄 Official NCERT Textbook PDF

Topic 6.2: Rate of Change of Quantities (Chain Rule in Action)

Drop a pebble into a calm Alappuzha backwater pond! Circular water ripples spread outward: when the radius grows at a steady 5 cm/s, does the wet circle's Area grow steadily too? NO! As the ripple gets wider, its outer rim gets longer — so dA/dt = 2πr · (dr/dt) grows faster and faster! That connected speedometer is Related Rates!

WHAT IS dy/dx?Rate of change of y per unit change in x! (+ = increasing, − = decreasing)

CHAIN RATE LINKdydt = dydx · dxdt (connects two changing quantities via time t!)

ECONOMICS MARGINALMarginal Cost = C'(x)
Marginal Revenue = R'(x)

🧠 Memory Trick: Differentiate FIRST, substitute numbers LAST! Plugging in r = 8 too early turns everything into a constant whose derivative is zero!

💡 Simple Kid-Check: Whenever a problem asks "How fast is Area/Volume/Height changing at the instant r = 5?", write the geometry formula first, take d/dt on both sides using Chain Rule, and ONLY AT THE END drop in r = 5!

3 Most Tested Geometry Rates:
• Circle Area A = πr² ⇒ dAdt = 2πr drdt | Sphere Vol V = 43πr³ ⇒ dVdt = 4πr² drdt
• Ladder against wall (x² + y² = L²) ⇒ 2xdxdt + 2ydydt = 0 (constant L gives 0!)

🪜 Sliding Ladder Example: A 5 m ladder leans against a wall. Its foot is pulled away at 2 cm/s. How fast does the top drop when x = 4 m?
• By Pythagoras at x = 4: 4² + y² = 5² ⇒ y = 3 m.
• Differentiate x² + y² = 25 wrt t: 2x(dx/dt) + 2y(dy/dt) = 0.
• Plug in x = 4, y = 3, dx/dt = +2 cm/s → 4(2) + 3(dy/dt) = 0 ⇒ dy/dt = −8/3 cm/s (height drops at 8/3 cm/s!).

Watch TWO classic board traps! (1) Units & Sign Trap: "Volume is decreasing at 8 cm³/s" means dV/dt = −8 (NEGATIVE!), and when writing your final English sentence ("decreasing at 8/3 cm/s"), drop the minus sign because the word "decreasing" already carries it! (2) Notice "rate of change wrt x" means d/dx (NO time t!), whereas "how fast / per second" means d/dt!

Topic 6.3: Increasing & Decreasing Functions via Sign of f'(x)

Imagine riding a roller-coaster at Wonderla Kochi! When the track climbs uphill from left to right, every tangent line slopes upward (f'(x) > 0 → Strictly Increasing). When it plunges downhill, tangent lines slope downward (f'(x) < 0 → Strictly Decreasing). Just checking the + or − sign of f'(x) instantly maps every climb and drop!

STRICTLY INCREASINGf'(x) > 0 on (a, b)
(Graph climbs left → right: x₁ < x₂ ⇒ f(x₁) < f(x₂))

STRICTLY DECREASINGf'(x) < 0 on (a, b)
(Graph drops left → right: x₁ < x₂ ⇒ f(x₁) > f(x₂))

WAVY SIGN CHART1) Solve f'(x) = 0
2) Split ℝ into intervals
3) Test sign of f'(x) in each!

🧠 Memory Trick: Think of walking LEFT to RIGHT along the curve: + slope = Uphill Climb (Increasing ↗), − slope = Downhill Slide (Decreasing ↘)!

💡 Simple Kid-Check: You don't even have to draw the graph to know where it rises or falls! Just differentiate, factor f'(x), mark the zeros on a number line, and test one easy number in each interval!

1
Find f'(x) & factor it cleanly. Solve f'(x) = 0 to get turning points.
2
Mark points on the number line to split ℝ into open disjoint intervals.
3
Check sign of f'(x): Intervals with f'(x) > 0 are Strictly Increasing; intervals with f'(x) < 0 are Strictly Decreasing!

🎢 Classic Interval Example: Find intervals where f(x) = 2x³ − 3x² − 36x + 7 is strictly increasing or decreasing.
• f'(x) = 6x² − 6x − 36 = 6(x² − x − 6) = 6(x − 3)(x + 2).
• Zeros at x = −2, 3 split ℝ into (−∞, −2), (−2, 3), (3, ∞).
• Test signs: + on (−∞, −2), − on (−2, 3), + on (3, ∞).
• Strictly Increasing: (−∞, −2) ∪ (3, ∞) | Strictly Decreasing: (−2, 3)!

Watch TWO sneaky traps! (1) On f(x) = x³, f'(x) = 3x² ≥ 0 (equals zero at one isolated point x = 0, not an entire flat interval) — so x³ is STILL strictly increasing on all of ℝ! (2) NEVER write "increasing on (−∞, −2) ∪ (3, ∞)" with a single union carelessly if f isn't monotonic across the jump — better yet, write "(−∞, −2) and (3, ∞)" separately (both notations appear in Kerala keys, listing them clearly avoids penalty)!

Topic 6.4: Critical Points, Local Maxima & Local Minima

Look at Kerala's Western Ghats mountain skyline — Idukki has local hilltops (Local Maxima) and lush valleys (Local Minima)! Notice something magical right at the exact summit of a smooth hill or the exact bottom of a valley: a tiny marble placed there stays flat because the tangent slope is horizontal (f'(c) = 0)! Those special candidate spots (plus sharp V-corners where f'(c) doesn't exist!) are called Critical Points!

CRITICAL POINT cInterior point where EITHER f'(c) = 0 OR f'(c) does NOT exist (sharp corner!)

LOCAL MAX / MINHighest / lowest point in its immediate neighbourhood (a valley can be lower than one hilltop yet higher than another!)

POINT OF INFLECTIONFlat point where f'(c) = 0, yet curve keeps climbing (e.g. x³ at 0) — NEITHER Max nor Min!

🧠 Memory Trick: Every smooth local extremum has f'(c) = 0, BUT f'(c) = 0 doesn't guarantee Max or Min (think of flat step x³ at x = 0)!

💡 Simple Kid-Check: "Local" simply means "king of your own street!" A local hilltop in Wayanad is higher than everything right around it, even if Anamudi peak far away is taller!

1
Stationary Point: Where slope becomes zero (f'(c) = 0) — could be a Peak, a Valley, or an Inflection shelf!
2
Non-Differentiable Corner: Functions like f(x) = |x − 3| + 2 still hit a sharp minimum 2 at x = 3 even though f'(3) doesn't exist!
3
Inflection Trap Check: On f(x) = x³, f'(0) = 0, yet left of 0 is negative and right of 0 is positive — it pauses flat for an instant and keeps climbing!

🏔️ Quick Critical Point Check: Find critical points of f(x) = x³ − 3x + 2.
• f'(x) = 3x² − 3 = 3(x − 1)(x + 1) = 0 ⇒ x = −1 (Local Max = 4) and x = +1 (Local Min = 0).

Watch TWO classic 1-mark conceptual traps! (1) Does a function with zero critical points like eˣ or log x have local maxima/minima? NO! (2) Can a Local Minimum value be GREATER than a Local Maximum value on the same graph? YES! Look at f(x) = x + 1x: Local Max at x = −1 is −2, while Local Min at x = +1 is +2!

Topic 6.4: First Derivative Test vs Second Derivative Test

Once you spot a suspect point where f'(c) = 0, how do you tell a Hilltop (Max) from a Valley (Min) without drawing the whole graph? You have TWO detective tools! First Derivative Test watches whether slope flips + → − (up then down = Hilltop!) or − → + (down then up = Valley!). Second Derivative Test checks the curve's curvature f''(c) — frown ∩ (f'' < 0) is Max, smile ∪ (f'' > 0) is Min!

1ST DERIVATIVE TESTAt c, check sign of f'(x):
+ → − ⇒ Local Max
− → + ⇒ Local Min
No change ⇒ Inflection!

2ND DERIVATIVE TESTIf f'(c) = 0:
f''(c) < 0 ⇒ Local Max (∩)
f''(c) > 0 ⇒ Local Min (∪)

IF f''(c) = 0?2nd Derivative Test is INCONCLUSIVE! Fall back immediately to 1st Derivative Test!

🧠 Memory Trick: Watch the sign flip! f'' < 0 (Negative = Sad Frown ∩) sits at a PEAK (Max)! f'' > 0 (Positive = Happy Smile ∪) sits at a VALLEY (Min)!

💡 Simple Kid-Check: Students mix up the signs of f''(c) all the time! Just picture a face: Negative f'' < 0 makes a sad cap ∩ (highest point at top = Local Max); Positive f'' > 0 makes a happy cup ∪ (lowest point at bottom = Local Min)!

Test Used	Local Maximum Condition	Local Minimum Condition	Neither (Inflection)
1st Derivative Test	`f'(x)` changes `+ → −`	`f'(x)` changes `− → +`	`+ → +` or `− → −`
2nd Derivative Test	`f'(c) = 0` & `f''(c) < 0`	`f'(c) = 0` & `f''(c) > 0`	`f''(c) = 0` → Use 1st Test!

🔍 Fast 2nd-Derivative Example: Classify extrema of f(x) = x³ − 6x² + 9x + 15.
• f'(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3) = 0 ⇒ c = 1, 3.
• f''(x) = 6x − 12.
• At x = 1: f''(1) = −6 < 0 (∩) ⇒ Local Max f(1) = 19.
• At x = 3: f''(3) = +6 > 0 (∪) ⇒ Local Min f(3) = 15!

TWO DEADLY BOARD TRAPS: (1) Writing "f''(c) > 0 means Local Maximum" — backwards! Positive f'' is a valley (MINIMUM)! (2) Writing "f''(0) = 0 so it's a point of inflection" — WRONG! Look at f(x) = x⁴: f'(0) = 0 AND f''(0) = 0, yet x⁴ has a Local Minimum at 0 (1st derivative goes − → +)! When f''(c) = 0, the 2nd test simply FAILS and you MUST use the 1st derivative test!

Topic 6.4: Absolute Maxima & Minima on a Closed Interval [a, b]

Who is the tallest student in a closed classroom? It could be a student standing on a hill stage inside the room, OR someone standing right at the left door x = a or right wall x = b! Whenever a function runs on a closed interval [a, b], the overall champion (Absolute Maximum / Minimum) might sit at an interior critical point OR right on the boundary walls a and b!

3-STEP CLOSED TEST1) Find critical pts c inside (a, b)
2) List ALL values: f(a), f(c), f(b)
3) Biggest = Abs Max; Smallest = Abs Min!

NO 2ND DERIVATIVE NEEDED!For Absolute Max/Min on [a, b], you NEVER need f''(x) — just compare f values!

OPEN vs CLOSEDContinuous function on closed [a, b] ALWAYS attains both Abs Max & Abs Min!

🧠 Memory Trick: "Throw Critical Points + Both Wall Endpoints into one Pot, Evaluate f(x), Pick Highest & Lowest!" Zero f'' guesswork needed!

💡 Simple Kid-Check: Local Max/Min only cares about nearby points inside the curve; Absolute Max/Min on [a, b] is the absolute king and lowest point across the entire fenced yard including the fences x = a and x = b!

1
Differentiate and solve f'(x) = 0. Keep ONLY those critical points c lying INSIDE [a, b]!
2
Compute f(x) at every valid critical point AND at the two endpoints a and b.
3
The largest number in your list is the Absolute Maximum; the smallest number is the Absolute Minimum!

👑 Endpoint Champion Example: Find absolute max & min of f(x) = 2x³ − 15x² + 36x + 1 on [1, 5].
• f'(x) = 6x² − 30x + 36 = 6(x − 2)(x − 3) = 0 ⇒ c = 2, 3 (both inside [1, 5]!).
• Evaluate ALL four candidates: f(1) = 24, f(2) = 29, f(3) = 28, f(5) = 56!
• Absolute Max = 56 (at endpoint x = 5, beating local peak 29!) | Absolute Min = 24 (at endpoint x = 1!).

Watch TWO costly mistakes! (1) Including a root of f'(x) = 0 that sits OUTSIDE [a, b]! Always filter out points outside the interval first! (2) Assuming a local max is automatically the absolute max — remember endpoints f(a) and f(b) often climb higher than any interior hill, just like f(5) = 56 above!

Topic 6.4: Real-Life Optimisation Word Problems (Boxes, Cylinders & Fencing)

Suppose you run a bakery in Thrissur and have an 18 cm × 18 cm tin sheet! You cut equal little squares of side x from the four corners and fold the sides up to bake an open cake box. Cut tiny squares (x ≈ 0) → super shallow tray, tiny volume! Cut huge squares (x ≈ 9) → tiny floor, zero volume! Somewhere in between sits the Goldilocks Maximum Volume — and Calculus finds it effortlessly!

STEP 1: ONE VARIABLEWrite target quantity (V, A, P) and use given constraint to eliminate extra variable!

STEP 2: SOLVE f'(x) = 0Differentiate target wrt x, set = 0, reject impossible sizes (x ≤ 0)!

STEP 3: VERIFY SIGN!ALWAYS prove f''(x) < 0 (Max) or f''(x) > 0 (Min) for full marks!

🧠 Memory Trick: "Constraint → One Letter → Differentiate → f'' Proof!" Symmetry almost always wins: max rectangle area = Square; max cylinder/cone problems share clean ratios!

💡 Simple Kid-Check: Every 4-mark optimisation problem follows the exact same recipe: you want to maximize/minimize ONE quantity (like Volume or Cost), given ONE fixed rule (like fixed Perimeter or Surface Area). Use the fixed rule to get down to ONE variable, differentiate, and check f''!

1
Draw & Name Variables: e.g. side x of cut corner, or radius r & height h of cylinder.
2
Eliminate Extra Variable: Use given perimeter P or surface area S so target is V(x) or A(x) in ONE variable!
3
Critical Value + Mandatory 2nd Derivative Check: Solve V'(x) = 0, check V''(x) < 0 (Max) or > 0 (Min), then answer the exact question asked!

📦 4-Mark Open Box Problem: Square tin sheet of side 12 cm; cut corner squares of side x.
• Base = (12 − 2x) by (12 − 2x), Height = x (where 0 < x < 6).
• Volume V(x) = x(12 − 2x)² = 4x(6 − x)² = 4(x³ − 12x² + 36x).
• V'(x) = 4(3x² − 24x + 36) = 12(x − 2)(x − 6) = 0 ⇒ x = 2 cm (reject x = 6 since base vanishes!).
• V''(x) = 12(2x − 8) ⇒ V''(2) = −48 < 0 ✓ (Maximum!). Max Volume = 2 × 8² = 128 cm³!

Don't throw away 1 easy mark! (1) Students solve V'(x) = 0 and STOP without checking V''(x) < 0 (or > 0) — Kerala marking keys ALWAYS deduct 1 mark if you don't prove it's actually a Max/Min! (2) Check what the question asks at the end: did it ask for the dimensions x, y OR the maximum volume/area itself? (3) In inscribed cylinder/cone problems with sphere radius R, remember R is a CONSTANT — only r, h vary!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March • 1 Marks): Find the rate of change of the area of a circle with respect to its radius r when r = 3 cm.

A = πr² ⇒ dAdr = 2πr.
At r = 3 cm: dAdr = 2π(3) = 6π cm²/cm.

Q2 (Kerala HSE Say • 2 Marks): The total revenue from selling x units of a product is R(x) = 13x² + 26x + 15. Find the marginal revenue when x = 7.

Marginal Revenue = R'(x) = 26x + 26.
At x = 7: R'(7) = 26(7) + 26 = 182 + 26 = ₹ 208.

Q3 (Kerala HSE March • 2 Marks): Find the intervals in which f(x) = 2x³ − 3x² − 36x + 7 is (a) strictly increasing (b) strictly decreasing.

f'(x) = 6x² − 6x − 36 = 6(x − 3)(x + 2); critical points x = −2, 3.
• Strictly Increasing: (−∞, −2) and (3, ∞) (where f'(x) > 0).
• Strictly Decreasing: (−2, 3) (where f'(x) < 0).

Q4 (NCERT / DHSE Model • 2 Marks): Prove that f(x) = x − sin x is an increasing function on ℝ.

f'(x) = 1 − cos x.
Since −1 ≤ cos x ≤ 1 for all x ∈ ℝ, we have 1 − cos x ≥ 0 everywhere.
Hence f(x) is increasing on ℝ.

Q5 (Kerala HSE March • 3 Marks): Prove that y = 4 sin θ2 + cos θ − θ is an increasing function of θ in [0, π/2].

dydθ = (2+cos θ)(4 cos θ) − 4 sin θ(−sin θ)(2 + cos θ)² − 1
= 8 cos θ + 4 − (4 + 4 cos θ + cos²θ)(2 + cos θ)² = cos θ (4 − cos θ)(2 + cos θ)².
In [0, π/2], cos θ ≥ 0 and 4 − cos θ > 0, so dy/dθ ≥ 0. Verified!

Q6 (Kerala HSE March • 3 Marks): Find all points of local maxima and local minima of f(x) = x³ − 6x² + 9x + 15.

f'(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3) = 0 ⇒ x = 1, 3.
f''(x) = 6x − 12.
• At x = 1: f''(1) = −6 < 0 ⇒ Local Max = 19.
• At x = 3: f''(3) = +6 > 0 ⇒ Local Min = 15.

Q7 (Kerala HSE Say • 3 Marks): Find the absolute maximum and absolute minimum values of f(x) = sin x + cos x, x ∈ [0, π].

f'(x) = cos x − sin x = 0 ⇒ tan x = 1 ⇒ x = π/4 ∈ [0, π].
Evaluate candidates: f(0) = 1, f(π/4) = 1√2 + 1√2 = √2, f(π) = −1.
• Absolute Max = √2 (at π/4) | Absolute Min = −1 (at π).

Q8 (Kerala HSE March • 4 Marks): A square piece of tin of side 18 cm is made into an open box by cutting equal squares from each corner and folding up flaps. What side square should be cut so the box volume is maximum?

Let cut square side be x (0 < x < 9).
V(x) = x(18 − 2x)² = 4(x³ − 18x² + 81x).
V'(x) = 4(3x² − 36x + 81) = 12(x − 3)(x − 9) = 0 ⇒ x = 3 cm (x = 9 impossible).
V''(x) = 12(2x − 12) ⇒ V''(3) = −72 < 0 (Max!).
Cut side = 3 cm; Max Volume = 3 × 12² = 432 cm³.