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Differentiability & Methods Exercise Solutions | KalviGavani

TN Class 11 Maths Ch10 Differentiability and Differentiation exercise solutions — first principles, product rule, quotient rule, chain rule with worked examples.

Updated 15 June 202610 min read

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1. How to Use This Page

This page contains complete exercise solutions for Chapter 10 — Differential Calculus: Differentiability and Methods of Differentiation of the Tamil Nadu State Board Class 11 Mathematics textbook. Solve each problem yourself first, then verify with these detailed solutions. Every worked example includes full step-by-step calculations.

2. Key Concepts (Quick Recap)

Derivative Rules

RuleFormula
Power ruled/dx(xⁿ) = n·x^(n−1)
Constantd/dx(c) = 0
Sum/Differenced/dx[f ± g] = f' ± g'
Product ruled/dx(uv) = u'v + uv'
Quotient ruled/dx(u/v) = (u'v − uv')/v²
Chain ruled/dx[f(g(x))] = f'(g(x)) · g'(x)

Standard Derivatives

d/dx(sin x) = cos x d/dx(cos x) = −sin x d/dx(tan x) = sec²x d/dx(cot x) = −csc²x d/dx(sec x) = sec x tan x d/dx(csc x) = −csc x cot x d/dx(eˣ) = eˣ d/dx(ln x) = 1/x d/dx(aˣ) = aˣ ln a d/dx(logₐx) = 1/(x ln a) d/dx(√x) = 1/(2√x) d/dx(1/x) = −1/x²

First Principles Definition

f'(x) = lim(h→0) [f(x + h) − f(x)] / h

Differentiability vs Continuity

  • Differentiable ⟹ Continuous (always true)
  • Continuous ⟹ Differentiable (NOT always true — e.g., f(x) = |x| at x = 0)

3. I. Choose the Best Answer (MCQs)

Q1. If f(x) = x³ + 2x, then f'(1) = ___

(a) 3   (b) 5   (c) 7   (d) 1

Answer(b) 5
Explanation
f(x) = x³ + 2x f'(x) = 3x² + 2 f'(1) = 3(1) + 2 = 5

Q2. d/dx(sin²x) equals:

(a) 2 sin x   (b) cos²x   (c) sin 2x   (d) −cos 2x

Answer(c) sin 2x
Explanation
d/dx(sin²x) = 2 sin x · d/dx(sin x) = 2 sin x cos x = sin 2x

Q3. If y = e^(3x), then dy/dx is:

(a) e^(3x)   (b) 3e^(3x)   (c) 3e^x   (d) e^(3x)/3

Answer(b) 3e^(3x)
ExplanationChain rule: dy/dx = e^(3x) · 3 = 3e^(3x)
Q4. The derivative of f(x) =xat x = 0 is:

(a) 1   (b) −1   (c) 0   (d) Does not exist

Answer(d) Does not exist
Explanation
LHD = lim(h→0⁻) [|0 + h| − |0|]/h = lim(h→0⁻) |h|/h = lim(h→0⁻)(−h/h) = −1 RHD = lim(h→0⁺) [|0 + h| − |0|]/h = lim(h→0⁺) |h|/h = lim(h→0⁺)(h/h) = 1 Since LHD ≠ RHD, f'(0) does not exist.

Q5. If y = x² · sin x, then dy/dx equals:

(a) 2x sin x   (b) x² cos x   (c) 2x sin x + x² cos x   (d) 2x cos x + x² sin x

Answer(c) 2x sin x + x² cos x
ExplanationProduct rule: u = x², v = sin x, dy/dx = 2x·sin x + x²·cos x

4. II. Fill in the Blanks

1. d/dx(5x⁴ − 3x² + 7) = 20x³ − 6x.

2. If y = cos(3x), then dy/dx = −3 sin(3x).

3. d/dx(eˣ²) = 2x · eˣ² (by chain rule).

4. The derivative of a constant function is 0.

5. d/dx(tan x) = sec²x.

6. If y = ln(5x), then dy/dx = 1/x.

7. d/dx(√x) = 1/(2√x).

8. For y = x⁵/x² = x³, dy/dx = 3x².

9. d/dx(1/x) = −1/x².

10. If f'(a) exists, then f is differentiable at x = a.

11. d/dx(xⁿ) evaluated at x = 1 gives n.

12. d/dx(sin x · cos x) = cos 2x (since sin x cos x = sin(2x)/2).

5. III. True or False (with Corrections)

1. Statement: A continuous function is always differentiable.

Answer: False. f(x) =xis continuous at x = 0 but not differentiable there. Differentiability implies continuity, but not vice versa.

2. Statement: The derivative of f(x) = eˣ is eˣ. Answer: True. eˣ is its own derivative.

3. Statement: In the product rule, (uv)' = u'v' + uv. Answer: False. (uv)' = u'v + uv'. Each function is differentiated once, not squared.

4. Statement: d/dx[f(g(x))] = f'(x) · g'(x). Answer: False. d/dx[f(g(x))] = f'(g(x)) · g'(x). The outer derivative is evaluated at g(x), not at x.

5. Statement: The derivative of f(x) = x at every point is 1. Answer: True. f(x) = x ⟹ f'(x) = 1 for all x.

6. IV. Match the Following

Column AColumn B
1. d/dx(sin x)a. sec²x
2. d/dx(cos x)b. 1/x
3. d/dx(tan x)c. cos x
4. d/dx(ln x)d. −sin x
5. d/dx(eˣ)e. eˣ
6. d/dx(sec x)f. sec x tan x
Answer1→c, 2→d, 3→a, 4→b, 5→e, 6→f

7. V. Answer Briefly (2-Mark Questions)

Q1. Differentiate f(x) = (3x + 5)⁴ using the chain rule.

Let u = 3x + 5, so f = u⁴ du/dx = 3 df/du = 4u³ = 4(3x + 5)³ df/dx = 4(3x + 5)³ × 3 = 12(3x + 5)³

Q2. Find dy/dx if y = (2x − 1)/(3x + 4) using the quotient rule.

u = 2x − 1, v = 3x + 4 u' = 2, v' = 3 dy/dx = (u'v − uv')/v² = (2(3x + 4) − (2x − 1)(3))/(3x + 4)² = (6x + 8 − 6x + 3)/(3x + 4)² = 11/(3x + 4)²

Q3. Differentiate y = x³ · eˣ.

Product rule: u = x³, v = eˣ u' = 3x², v' = eˣ dy/dx = u'v + uv' = 3x² · eˣ + x³ · eˣ = eˣ(x³ + 3x²)

Q4. Differentiate y = sin(2x + 3).

Chain rule: let u = 2x + 3 dy/dx = cos(2x + 3) · d/dx(2x + 3) = 2 cos(2x + 3)

8. VI. Answer in Detail (5-Mark Questions)

Q1. Differentiate f(x) = x² + 3x − 5 from first principles.

Step 1: Write the definition f'(x) = lim(h→0) [f(x+h) − f(x)]/h Step 2: Find f(x+h) f(x+h) = (x+h)² + 3(x+h) − 5 = x² + 2xh + h² + 3x + 3h − 5 Step 3: Find f(x+h) − f(x) = (x² + 2xh + h² + 3x + 3h − 5) − (x² + 3x − 5) = 2xh + h² + 3h Step 4: Divide by h [f(x+h) − f(x)]/h = (2xh + h² + 3h)/h = 2x + h + 3 Step 5: Take the limit as h → 0 f'(x) = lim(h→0)(2x + h + 3) = 2x + 3 Verification: d/dx(x² + 3x − 5) = 2x + 3 ✓

Q2. Find dy/dx if y = (x² + 1)/(x² − 1) using the quotient rule. Simplify the result.

Step 1: Identify u and v u = x² + 1, v = x² − 1 u' = 2x, v' = 2x Step 2: Apply quotient rule dy/dx = (u'v − uv')/v² = (2x(x² − 1) − (x² + 1)(2x))/(x² − 1)² Step 3: Expand numerator = (2x³ − 2x − 2x³ − 2x)/(x² − 1)² = (−4x)/(x² − 1)² Therefore dy/dx = −4x/(x² − 1)²

Q3. Differentiate y = sin²x + cos²x + tan²x from first principles (for the first two terms) and by rules (for the third term).

Step 1: Simplify using identity sin²x + cos²x = 1 (Pythagorean identity) So y = 1 + tan²x Step 2: Use identity for tan²x y = sec²x (since 1 + tan²x = sec²x) Step 3: Differentiate dy/dx = d/dx(sec²x) = 2 sec x · sec x tan x = 2 sec²x tan x Alternative approach — differentiating the original expression: dy/dx = 2 sin x cos x + 2 cos x(−sin x) + 2 tan x sec²x = 2 sin x cos x − 2 sin x cos x + 2 tan x sec²x = 2 tan x sec²x Both methods give the same answer ✓

Q4. Find dy/dx if y = √((1 − sin x)/(1 + sin x)) using the chain rule and quotient rule.

Step 1: Let y = √u where u = (1 − sin x)/(1 + sin x) Step 2: Chain rule dy/dx = (1/2√u) · du/dx Step 3: Find du/dx using quotient rule u = (1 − sin x)/(1 + sin x) Let p = 1 − sin x, q = 1 + sin x p' = −cos x, q' = cos x du/dx = (p'q − pq')/q² = (−cos x(1 + sin x) − (1 − sin x)(cos x))/(1 + sin x)² = (−cos x − cos x sin x − cos x + cos x sin x)/(1 + sin x)² = (−2 cos x)/(1 + sin x)² Step 4: Substitute back dy/dx = (1/2√u) × (−2 cos x)/(1 + sin x)² = (−cos x)/[(1 + sin x)² · √((1 − sin x)/(1 + sin x))] Step 5: Simplify further Multiply numerator and denominator by √(1 + sin x): = (−cos x)/[(1 + sin x)^(3/2) · √(1 − sin x)] Note: Using the identity √(1−sin²x) = |cos x| = cos x (for x in appropriate range): The answer simplifies to −cos x/((1 + sin x)√(cos²x)) = −cos x/((1 + sin x)|cos x|) For cos x > 0: dy/dx = −1/(1 + sin x) · cos x/cos x · 1/√(1−sin²x/(1+sin x)²)

Q5. If y = (3x + 2)⁵ · (4x − 1)³, find dy/dx using the product rule with chain rule.

Step 1: Identify the two factors u = (3x + 2)⁵, v = (4x − 1)³ Step 2: Find u' and v' using chain rule u' = 5(3x + 2)⁴ · 3 = 15(3x + 2)⁴ v' = 3(4x − 1)² · 4 = 12(4x − 1)² Step 3: Apply product rule dy/dx = u'v + uv' = 15(3x + 2)⁴ · (4x − 1)³ + (3x + 2)⁵ · 12(4x − 1)² Step 4: Factor out common terms = 3(3x + 2)⁴(4x − 1)² [5(4x − 1) + 4(3x + 2)] = 3(3x + 2)⁴(4x − 1)² [20x − 5 + 12x + 8] = 3(3x + 2)⁴(4x − 1)²(32x + 3) Therefore: dy/dx = 3(3x + 2)⁴(4x − 1)²(32x + 3)

9. Common Mistakes to Avoid

1. Forgetting the inner derivative in chain rule: If y = sin(3x), then dy/dx = cos(3x) · 3, NOT just cos(3x). Always multiply by the derivative of the inner function.

2. Sign error in quotient rule: The formula is (u'v − uv')/v², with a MINUS sign. Many students write u'v + uv' by mistake.

3. Product rule is NOT (uv)' = u'v': The correct formula is (uv)' = u'v + uv'. You need both terms.

4. Confusing d/dx(eˣ) with d/dx(aˣ): d/dx(eˣ) = eˣ but d/dx(aˣ) = aˣ ln a. Don't forget the ln a factor for general exponential bases.

5. Differentiability vs continuity: Differentiable ⟹ Continuous, but NOT the other way around.xis continuous but not differentiable at x = 0.

6. First principles algebra errors: When expanding f(x+h), be careful with signs and exponents. Double-check each step before taking the limit.

10. Exam Checklist

  • [ ] Can differentiate from first principles (limit definition)
  • [ ] Know power rule: d/dx(xⁿ) = n·x^(n−1)
  • [ ] Can apply product rule: (uv)' = u'v + uv'
  • [ ] Can apply quotient rule: (u/v)' = (u'v − uv')/v²
  • [ ] Can apply chain rule for composite functions
  • [ ] Know all standard derivatives (trig, exponential, logarithmic)
  • [ ] Can differentiate implicit functions
  • [ ] Understand the relationship: differentiable ⟹ continuous (but not vice versa)
  • [ ] Can find derivatives of parametric functions: dy/dx = (dy/dt)/(dx/dt)
  • [ ] Can simplify derivative expressions by factoring common terms

Frequently Asked Questions

If y = f(g(x)), then dy/dx = f'(g(x)) · g'(x). Differentiate the outer function, then multiply by the derivative of the inner function.