NCERT Plus Two (Class 12) Mathematics • Chapter 5

Continuity and Differentiability

Q: Find dy dx if y = cos(√x) ( x > 0 ).

dy dx = −sin(√x) · d dx (√x) = − sin(√x) 2√x .

Q: Check whether f(x) = |x| is differentiable at x = 0 .

LHD = lim h→0 |0−h| − 0 −h = −1 . RHD = lim h→0 |0+h| − 0 h = +1 . Since LHD ≠ RHD , |x| is NOT differentiable at x = 0 .

Q: Find k if f(x) = { k cos x π − 2x if x ≠ π/2; 3 if x = π/2 } is continuous at x = π/2 .

Put x = π/2 − h (as x → π/2, h → 0 ): lim x→π/2 k cos x π − 2x = lim h→0 k sin h 2h = k 2 · 1 = k 2 . For continuity, k 2 = f(π/2) = 3 ⇒ k = 6 .

Q: Find dy dx if y + sin y = cos x .

Differentiating both sides wrt x : dy dx + cos y · dy dx = −sin x ⇒ (1 + cos y) dy dx = −sin x ⇒ dy dx = − sin x 1 + cos y ( y ≠ (2n+1)π ).

Q: Differentiate y = tan⁻¹( cos x − sin x cos x + sin x ) , 0 < x < π , with respect to x .

Divide numerator & denominator inside by cos x : y = tan⁻¹( 1 − tan x 1 + tan x ) = tan⁻¹(tan( π 4 − x)) = π 4 − x . Therefore dy dx = −1 !

Q: Find dy dx if y = x^(sin x) + (sin x)^(cos x) .

Let u = x^(sin x) and v = (sin x)^(cos x) , so dy dx = du dx + dv dx . • log u = sin x log x ⇒ du dx = x^(sin x)[ sin x x + cos x log x] . • log v = cos x log(sin x) ⇒ dv dx = (sin x)^(cos x)[cos x cot x − sin x log(sin x)] . • Sum both pieces for final dy dx .

Q: If x = a(θ − sin θ) and y = a(1 − cos θ) , find dy dx .

dx dθ = a(1 − cos θ) and dy dθ = a sin θ . dy dx = a sin θ a(1 − cos θ) = 2 sin(θ/2) cos(θ/2) 2 sin²(θ/2) = cot( θ 2 ) .

Q: If y = (tan⁻¹ x)² , show that (x² + 1)² y₂ + 2x(x² + 1) y₁ = 2 .

1. y₁ = 2 tan⁻¹ x 1 + x² ⇒ (1 + x²) y₁ = 2 tan⁻¹ x . 2. Differentiate wrt x : (1 + x²) y₂ + 2x y₁ = 2 1 + x² . 3. Multiply throughout by (1 + x²) : (x² + 1)² y₂ + 2x(x² + 1) y₁ = 2 . Hence proved!

Continuity at a point, chain rule, implicit & inverse-trig derivatives, exponential/log rules, logarithmic & parametric differentiation, and second-order derivatives.

📄 Official NCERT Textbook PDF

Topic 5.2: Continuity at a Point & Algebra of Continuous Functions

Imagine cycling smoothly along NH-66 from Kochi to Thrissur without lifting your tyres off the tar! That unbroken road is a Continuous Function. If a canal bridge suddenly breaks and drops two metres (a Jump), or one manhole cover is missing and placed on the footpath (a Removable Hole), you cannot cross without jumping — the curve is Discontinuous right there!

3-WAY CONTINUITY TESTf(x) is continuous at x = c iff
LHL = RHL = f(c) (finite!)

ALWAYS CONTINUOUSPolynomials, sin x, cos x, eˣ, |x| & constants are continuous on all of ℝ!

ALGEBRA RULEIf f, g continuous at c, so are f ± g, f·g, f/g (g(c) ≠ 0) & f(g(x))

🧠 Memory Trick: Think of the 3-Lock Bridge — Left approach road (LHL), Right approach road (RHL), and the middle pillar (f(c)) must ALL meet at one exact height!

💡 Simple Kid-Check: Can you trace the graph past x = c in one single stroke without lifting your pen? If YES, it's continuous at c! Even a sharp V-bend like |x| is continuous because you never lift your pen!

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Left-Hand Limit (LHL): Walk toward c from the left: lim_h→0 f(c − h).
2
Right-Hand Limit (RHL): Walk toward c from the right: lim_h→0 f(c + h).
3
Exact Value f(c): Check the solid dot at x = c. Continuity holds ONLY if LHL = RHL = f(c)!

Key Limit Aid: lim_x→0 sin kxx = k | lim_x→0 1 − cos xx² = 12 | lim_x→0 eˣ − 1x = 1

🧒 Super-Easy Example: Find k if f(x) = kx + 1 (for x ≤ 5) and 3x − 5 (for x > 5) is continuous at x = 5.
• LHL = f(5) = 5k + 1 | RHL = 3(5) − 5 = 10.
• Set LHL = RHL → 5k + 1 = 10 ⇒ 5k = 9 ⇒ k = 9/5!

Watch two classic DHSE traps! (1) Don't stop after checking LHL = RHL — a hole like (x² − 4)/(x − 2) with f(2) = 1 has LHL = RHL = 4 ≠ f(2), so it is STILL discontinuous! (2) f(x) = |x| IS continuous at 0 (`LHL = RHL = f(0) = 0`), and tan x is continuous on its domain (only undefined at odd multiples of π/2).

Topic 5.3: Differentiability & Chain Rule for Composite Functions

Look at a sharp coconut-palm leaf roof tip shaped like V (|x|). You can slide your finger across the tip without lifting it (continuous!), yet can you balance one flat ruler smoothly right on the pointed peak? No! The left slope (−1) and right slope (+1) clash — so a sharp corner is NOT differentiable! Meanwhile, nested functions behave like linked bicycle gears using the Chain Rule!

DIFFERENTIABLE AT cLHD = RHD (both finite!).
Differentiable ⇒ Continuous (Converse FALSE!)

CHAIN RULEddx[f(g(x))] = f'(g(x)) · g'(x)

FAST ROOTS(√u)' = u'2√u | (uⁿ)' = n·uⁿ⁻¹·u'

🧠 Memory Trick: Onion-Peel Rule! Peel the OUTER layer first while keeping the inside frozen, then multiply by the derivative of the INNER layer until you reach x!

💡 Simple Kid-Check: Differentiability means "smooth with no sharp kinks or vertical cliffs". If a pedal gear spins twice as fast as the pedal (u = 2x) and the back wheel spins at rate cos u, the total speed is simply the product: 2 · cos(2x)!

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Left & Right Derivatives: LHD = lim_h→0 f(c−h) − f(c)−h and RHD = lim_h→0 f(c+h) − f(c)h. Corners give LHD ≠ RHD!
2
Chain Rule 3-Layer Stack: If y = f(g(h(x))), differentiate outside-in: f'(g(h(x))) · g'(h(x)) · h'(x).

⚙️ Gear-Chain Example: Differentiate y = sin(cos(x²)):
• Outer sin(□) → cos(cos(x²))
• Middle cos(○) → · (−sin(x²))
• Inner x² → · (2x)
• Final Answer = −2x sin(x²) cos(cos(x²))!

Two classic errors cost easy marks! (1) Writing d/dx [sin(x²)] = cos(2x) — WRONG! Keep x² frozen inside cos(x²) and multiply OUTSIDE by 2x! (2) Assuming every continuous function is differentiable: remember |x − a| is continuous at x = a, yet NEVER differentiable at x = a (corner!). Also [x] (greatest integer) is neither continuous nor differentiable at integers!

Topic 5.3: Implicit Functions & Inverse Trigonometric Derivatives

Sometimes x and y get tangled together like earphones in your pocket — e.g. x² + y² = 25 or x + sin(xy) = y! You don't have to untangle y = ... first! Differentiate BOTH sides straight away, and whenever you touch a y, just attach its buddy dydx by Chain Rule!

IMPLICIT RULEDifferentiate every term wrt x.
Any y-term gets · dydx! Then group dydx!

TOP 3 INV TRIG(sin⁻¹x)' = 1√(1−x²) | (tan⁻¹x)' = 11+x²
(cos⁻¹x)' = −1√(1−x²)

CO-PARTNER RULE(cot⁻¹x)' = −11+x²
(sec⁻¹x)' = 1|x|√(x²−1) (cosec⁻¹ is −!)

🧠 Memory Trick: Every "co-" inverse function (cos⁻¹, cot⁻¹, cosec⁻¹) has the EXACT same derivative as its partner, just wearing a MINUS sign!

💡 Simple Kid-Check: For implicit equations like y², pretend y is a mystery box containing x: Chain Rule gives 2y · (dy/dx)! Collect all dy/dx pieces on the left and factor it out cleanly!

Function `f(x)`	Derivative `f'(x)`	Valid Domain
`sin⁻¹ x` / `cos⁻¹ x`	`± 1√(1 − x²)`	`x ∈ (−1, 1)`
`tan⁻¹ x` / `cot⁻¹ x`	`± 11 + x²`	`x ∈ ℝ`
`sec⁻¹ x` / `cosec⁻¹ x`	`± 1\|x\|√(x² − 1)`	`\|x\| > 1`

🔎 Quick Implicit Example: Find dydx if x² + xy + y² = 100.
• Differentiate term-by-term (use Product Rule on xy!): 2x + [x·dydx + y·1] + 2y·dydx = 0
• Group dy/dx: (x + 2y)dydx = −(2x + y) ⇒ dydx = −2x + yx + 2y!

Watch two classic mistakes! (1) On terms like xy or sin(xy), students forget the Product Rule: d/dx(xy) = x(dy/dx) + y! (2) For messy inverse trig like y = tan⁻¹(2x1−x²), NEVER differentiate straight with the monster quotient rule! Substitute x = tan θ first to simplify to y = 2 tan⁻¹ x → instant 21+x²!

Topic 5.4: Exponential (`eˣ`) & Natural Logarithmic (`log x`) Derivatives

Imagine a magical money tree in Thrissur whose growth speed at every instant equals its exact height: when it is 2 m tall, it grows at 2 m/hour; at 10 m, 10 m/hour! Out of all numbers in mathematics, ONLY e ≈ 2.71828 makes the curve's slope equal its height everywhere: (eˣ)' = eˣ! Meanwhile, its mirror twin logₑ x flattens out with slope 1/x!

SELF-CLONE eˣddx(eˣ) = eˣ | (eᵘ)' = eᵘ · u'

NATURAL LOGddx(log x) = 1x | [log u]' = u'u

GENERAL BASE aˣddx(aˣ) = aˣ logₑ a (a > 0)

🧠 Memory Trick: eˣ is the Teflon superhero — differentiate it 100 times, still eˣ! And log(blob) always flips the blob into (blob') / (blob)!

💡 Simple Kid-Check: In NCERT Plus Two, plain log x ALWAYS means natural log to base e (often written ln x on calculators). Because y = eˣ and y = log x are reflections across the diagonal mirror y = x, their slopes are reciprocals!

3 Log Laws That Save Lives Before Differentiating:
• log(AB) = log A + log B | log(A/B) = log A − log B | log(Aⁿ) = n log A

⚡ Fast Chain-Rule Examples:
1. d/dx [e^(−3x)] = −3e^(−3x) | d/dx [e^(cos x)] = −sin x · e^(cos x)
2. d/dx [log(cos eˣ)] = (cos eˣ)'cos eˣ = −sin(eˣ) · eˣcos(eˣ) = −eˣ tan(eˣ)!

Watch this massive trap! What is d/dx (2ˣ)? Students write x · 2^(x−1) — WRONG! The Power Rule works ONLY when the exponent is constant (x²)! For 2ˣ (constant base, variable power), the correct answer is 2ˣ logₑ 2! Also, if a question asks d/dx (log₁₀ x), convert base first: logₑ xlogₑ 10 → derivative 1x logₑ 10!

Topic 5.5: Logarithmic Differentiation for u(x)^v(x) & Product Chains

What happens when BOTH the base AND the exponent vary — like y = xˣ or y = (sin x)^(cos x)? Power Rule shouts "The exponent isn't constant!", and Exponential Rule shouts "The base isn't constant!" How do we rescue it? Pull the Log Zipper! Taking log on both sides pulls that stubborn floating exponent right down to the floor as simple multiplication!

WHEN TO USE LOGS1) Variable^Variable like u(x)^v(x)
2) Giant 3+ factor products/roots

3-STEP LOG ZIPPER1) log y = v log u
2) Differentiate: 1ydydx = ...
3) Multiply right side by y!

ICONIC RESULTddx(xˣ) = xˣ(1 + log x)

🧠 Memory Trick: Never log a PLUS! For y = xˣ + (sin x)ˣ, split y = u + v FIRST, take logs on u and v separately, then add du/dx + dv/dx!

💡 Simple Kid-Check: Compare the three look-alikes carefully: x² (constant power → Power Rule 2x), 2ˣ (constant base → Exp Rule 2ˣ log 2), and xˣ (BOTH vary → Log Zipper xˣ(1 + log x))!

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Take log on both sides: log y = x log x. (Power drops down cleanly!)
2
Differentiate implicitly wrt x: 1ydydx = x·1x + (log x)·1 = 1 + log x.
3
Multiply by y: dydx = xˣ(1 + log x)!

🏆 Classic 4-Mark Board Problem: Differentiate y = x^(sin x) + (sin x)^(cos x).
• Let u = x^(sin x), v = (sin x)^(cos x), so y = u + v ⇒ dydx = dudx + dvdx.
• log u = sin x · log x ⇒ dudx = x^(sin x)[sin xx + cos x log x].
• log v = cos x · log(sin x) ⇒ dvdx = (sin x)^(cos x)[cos x cot x − sin x log(sin x)].
• Add them up — full 4/4 marks guaranteed!

#1 Kerala HSE Mistake: When given a SUM like y = xˣ + aˣ or x^(sin x) + (sin x)^(cos x), students take logs directly and write log y = log(u) + log(v) — DEAD WRONG because log(u + v) ≠ log u + log v! Always split into y = u + v first! Also remember when differentiating y = u·v, don't forget to multiply by y at the final step!

Topic 5.6 – 5.7: Parametric Differentiation & Second-Order Derivatives (`d²y/dx²`)

Track a cricket ball flying across Greenfield Stadium: at any second t, its forward distance is x = f(t) and height is y = g(t)! Here t is the invisible timekeeper (Parameter). To find the path slope dy/dx without deleting t, just divide vertical speed dy/dt by horizontal speed dx/dt! Differentiate once more wrt x to get the Second-Order Derivative d²y/dx²!

PARAMETRIC dy/dxdydx = dy/dtdx/dt (dx/dt ≠ 0)

2ND ORDER d²y/dx²y₂ = ddx(dydx)
Parametric: d/dt (dy/dx)dx/dt!

SHOW-THAT TRICKCross-multiply denominators (like √(1−x²) or 1+x²) BEFORE taking 2nd derivative!

🧠 Memory Trick: Parametric 2nd derivative ALWAYS divides by dx/dt a SECOND time: d²y/dx² = [d/dt(y')] ÷ (dx/dt)!

💡 Simple Kid-Check: Notice the notation: first derivative is y' or y₁ or dy/dx; second derivative is y'' or y₂ or d²y/dx² — it simply measures how fast the slope itself bends (concavity)!

Classic Parametric Pairs:
• x = a t², y = 2at ⇒ dydx = 1t, d²ydx² = −1/t²2at = −12at³
• x = a(θ − sin θ), y = a(1 − cos θ) ⇒ dydx = sin θ1 − cos θ = cot(θ2)

🏆 Golden 4-Mark Proof Trick: Given y = (tan⁻¹ x)², show (x² + 1)² y₂ + 2x(x² + 1) y₁ = 2.
• Step 1: y₁ = 2 tan⁻¹ x · 11 + x².
• Step 2 (PRO MOVE — clear fraction FIRST!): (1 + x²) y₁ = 2 tan⁻¹ x.
• Step 3 (Differentiate again by Product Rule!): (1 + x²) y₂ + 2x y₁ = 21 + x².
• Multiply both sides by (1 + x²) → (x² + 1)² y₂ + 2x(x² + 1) y₁ = 2! Done in 4 lines!

Watch out for TWO famous pitfalls! (1) In parametric 2nd-order problems like x = a cos θ, y = b sin θ, finding d/dθ(dy/dx) is NOT d²y/dx² — you MUST divide once more by dx/dθ! (2) Remember d²y/dx² ≠ (dy/dx)², and in "show that" proofs, always cross-multiply denominators like (1 + x²) or √(1 − x²) BEFORE differentiating the second time!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March • 1 Marks): Find dydx if y = cos(√x) (x > 0).

dydx = −sin(√x) · ddx(√x) = −sin(√x)2√x.

Q2 (Kerala HSE Say • 1 Marks): Check whether f(x) = |x| is differentiable at x = 0.

LHD = lim_h→0 |0−h| − 0−h = −1.
RHD = lim_h→0 |0+h| − 0h = +1.
Since LHD ≠ RHD, |x| is NOT differentiable at x = 0.

Q3 (Kerala HSE March • 2 Marks): Find k if f(x) = { k cos xπ − 2x if x ≠ π/2; 3 if x = π/2 } is continuous at x = π/2.

Put x = π/2 − h (as x → π/2, h → 0):
lim_x→π/2 k cos xπ − 2x = lim_h→0 k sin h2h = k2 · 1 = k2.
For continuity, k2 = f(π/2) = 3 ⇒ k = 6.

Q4 (Kerala HSE Model • 2 Marks): Find dydx if y + sin y = cos x.

Differentiating both sides wrt x:
dydx + cos y · dydx = −sin x
⇒ (1 + cos y)dydx = −sin x ⇒ dydx = −sin x1 + cos y (y ≠ (2n+1)π).

Q5 (NCERT / DHSE March • 3 Marks): Differentiate y = tan⁻¹(cos x − sin xcos x + sin x), 0 < x < π, with respect to x.

Divide numerator & denominator inside by cos x:
y = tan⁻¹(1 − tan x1 + tan x) = tan⁻¹(tan(π4 − x)) = π4 − x.
Therefore dydx = −1!

Q6 (Kerala HSE March • 3 Marks): Find dydx if y = x^(sin x) + (sin x)^(cos x).

Let u = x^(sin x) and v = (sin x)^(cos x), so dydx = dudx + dvdx.
• log u = sin x log x ⇒ dudx = x^(sin x)[sin xx + cos x log x].
• log v = cos x log(sin x) ⇒ dvdx = (sin x)^(cos x)[cos x cot x − sin x log(sin x)].
• Sum both pieces for final dydx.

Q7 (Kerala HSE Say • 3 Marks): If x = a(θ − sin θ) and y = a(1 − cos θ), find dydx.

dxdθ = a(1 − cos θ) and dydθ = a sin θ.
dydx = a sin θa(1 − cos θ) = 2 sin(θ/2) cos(θ/2)2 sin²(θ/2) = cot(θ2).

Q8 (Kerala HSE March • 4 Marks): If y = (tan⁻¹ x)², show that (x² + 1)² y₂ + 2x(x² + 1) y₁ = 2.

1. y₁ = 2 tan⁻¹ x1 + x² ⇒ (1 + x²) y₁ = 2 tan⁻¹ x.
2. Differentiate wrt x: (1 + x²) y₂ + 2x y₁ = 21 + x².
3. Multiply throughout by (1 + x²):
(x² + 1)² y₂ + 2x(x² + 1) y₁ = 2. Hence proved!