Posted by FUZY MATH ACADEMY September 26, 2025

Continuity and Differentiability – Class 12 Maths (Chapter 5)

class 12th student taking online class
Continuity and Differentiability – Class 12 Maths (Chapter 5)

Continuity and Differentiability is one of the most important chapters in Class 12 Mathematics. It builds upon the concepts of limits and introduces advanced techniques of differentiation, chain rule, logarithmic differentiation, exponential functions, and implicit functions. This chapter is not only crucial for board exams but also forms the foundation for competitive exams like JEE, NEET, and other entrance tests. At FUZY MATH ACADEMY, powered by an advanced LMS and 24/7 AI Math Solver support, students can learn Continuity and Differentiability through structured lessons, step-by-step solutions, and regular practice.

Class 12 student

Important Formulas of Continuity and Differentiability

Continuity

A function f(x) is continuous at x = a if:
lim (x → a⁻) f(x) = lim (x → a⁺) f(x) = f(a)

Differentiability

A function f(x) is differentiable at x = a if left-hand derivative (LHD) = right-hand derivative (RHD).
If f(x) is differentiable at a point, then it is continuous at that point (but not vice-versa).

Standard Derivatives

(d/dx)(xⁿ) = n·xⁿ⁻¹
(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)(aˣ) = aˣ ln a
(d/dx)(log x) = 1/x

Differentiation Rules

Product Rule: (uv)' = u'v + uv'
Quotient Rule: (u/v)' = (u'v – uv') / v²
Chain Rule: d/dx [f(g(x))] = f'(g(x))·g'(x)

Logarithmic Differentiation

If y = f(x) = uᵛ, then log y = v log u, then differentiate both sides.

Second Order Derivatives

If y = f(x), then d²y/dx² = derivative of (dy/dx).

Class 12 student

Q1–Q20 (Easy)

Q1: Evaluate lim_x→0 (sin x)/x.
Formula Used: lim_x→0 (sin x)/x = 1

Solution: By standard limit, value = 1.
Q2: Differentiate f(x) = x².
Formula Used: d/dx(xⁿ) = n·xⁿ⁻¹

Solution: f’(x) = 2x.
Q3: Differentiate f(x) = √x.
Formula: d/dx(√x) = 1/(2√x)

Solution: f’(x) = 1/(2√x).
Q4: Find slope of curve y = x³ at x = 2.
Formula: dy/dx = 3x²

Solution: dy/dx = 3(2²) = 12.
Q5: Differentiate sin x.
Formula: d/dx(sin x) = cos x

Solution: f’(x) = cos x.
Q6: Differentiate cos x.
Formula: d/dx(cos x) = -sin x

Solution: f’(x) = -sin x.
Q7: Differentiate tan x.
Formula: d/dx(tan x) = sec²x

Solution: f’(x) = sec²x.
Q8: Differentiate log x.
Formula: d/dx(log x) = 1/x

Solution: f’(x) = 1/x.
Q9: Differentiate eˣ.
Formula: d/dx(eˣ) = eˣ

Solution: f’(x) = eˣ.
Q10: Differentiate aˣ.
Formula: d/dx(aˣ) = aˣ ln a

Solution: f’(x) = aˣ ln a.
Q11: lim_x→0 (1 - cos x)/x².
Formula: lim (1 - cos x)/x² = 1/2

Solution: Value = 1/2.
Q12: Derivative of constant function.
Formula: d/dx(C) = 0

Solution: Answer = 0.
Q13: Differentiate f(x) = 3x + 5.
Formula: d/dx(ax+b) = a

Solution: f’(x) = 3.
Q14: Differentiate sec x.
Formula: d/dx(sec x) = sec x tan x

Solution: f’(x) = sec x tan x.
Q15: Differentiate cosec x.
Formula: d/dx(cosec x) = -cosec x cot x

Solution: f’(x) = -cosec x cot x.
Q16: Differentiate cot x.
Formula: d/dx(cot x) = -cosec²x

Solution: f’(x) = -cosec²x.
Q17: Find derivative of y = 5x⁴.
Formula: d/dx(xⁿ) = n·xⁿ⁻¹

Solution: y’ = 20x³.
Q18: Differentiate y = 1/x.
Formula: d/dx(x⁻¹) = -x⁻²

Solution: y’ = -1/x².
Q19: Differentiate f(x) = sin²x.
Formula: Chain Rule

Solution: f’(x) = 2 sin x cos x = sin 2x.
Q20: Differentiate cos²x.
Formula: Chain Rule

Solution: f’(x) = -2 cos x sin x = -sin 2x.

Q21–Q40 (Moderate)

Q21: Differentiate f(x) = sin x cos x.
Formula: Product Rule

Solution: f’(x) = cos²x - sin²x = cos 2x.
Q22: Differentiate y = x log x.
Formula: Product Rule

Solution: y’ = 1·log x + x·(1/x) = log x + 1.
Q23: Differentiate y = xᵉ.
Formula: Logarithmic Differentiation

Solution: y’ = e xᵉ⁻¹.
Q24: Differentiate y = (x² + 1)³.
Formula: Chain Rule

Solution: y’ = 3(x²+1)²·2x = 6x(x²+1)².
Q25: Differentiate y = ln(sin x).
Formula: d/dx ln u = u’/u

Solution: y’ = (cos x)/(sin x) = cot x.
Q26: Differentiate y = ln(cos x).
Solution: y’ = (-sin x)/(cos x) = -tan x.
Q27: Differentiate y = tan⁻¹(2x).
Formula: d/dx tan⁻¹ u = u’/(1+u²)

Solution: y’ = (2)/(1+4x²).
Q28: Differentiate y = sin⁻¹(x).
Formula: d/dx sin⁻¹x = 1/√(1-x²)

Solution: y’ = 1/√(1-x²).
Q29: Differentiate y = cos⁻¹x.
Solution: y’ = -1/√(1-x²).
Q30: Differentiate y = sec⁻¹x.
Formula: d/dx sec⁻¹x = 1/(|x|√(x²-1))

Solution: y’ = 1/(x√(x²-1)).
Q31: Differentiate y = xᵐ log x.
Solution: y’ = m xᵐ⁻¹ log x + xᵐ(1/x).
Q32: Differentiate y = eˣ sin x.
Formula: Product Rule

Solution: y’ = eˣ(sin x + cos x).
Q33: Differentiate y = eˣ cos x.
Solution: y’ = eˣ(cos x - sin x).
Q34: Differentiate y = eˣ tan x.
Solution: y’ = eˣ(tan x + sec²x).
Q35: Differentiate y = √(1+x²).
Solution: y’ = x/√(1+x²).
Q36: Differentiate y = 1/(1+x).
Solution: y’ = -1/(1+x)².
Q37: Differentiate y = log(x²+1).
Solution: y’ = 2x/(x²+1).
Q38: Differentiate y = tan⁻¹(√x).
Solution: y’ = 1/(2√x(1+x)).
Q39: Differentiate y = sin(x²).
Solution: y’ = 2x cos(x²).
Q40: Differentiate y = cos(x²).
Solution: y’ = -2x sin(x²).

Q41–Q60 (Higher NCERT Type)

Q41. Evaluate lim_x→0 (sin x − x) / x³.

Formula used: Maclaurin series for sin x = x − x³/6 + O(x⁵).
1. Expand numerator: sin x − x = (x − x³/6 + ... ) − x = −x³/6 + O(x⁵).
2. So (sin x − x)/x³ = (−x³/6 + ...)/x³ → −1/6 as x→0.
3. Answer: −1/6.
Q42. Let f(x)=x² sin(1/x) for x≠0 and f(0)=0. Show f is differentiable at 0.

Formula used: Definition of derivative: f'(0)=lim_h→0[f(h)−f(0)]/h; squeeze theorem.
1. Compute the difference quotient: [f(h)−f(0)]/h = [h² sin(1/h)]/h = h·sin(1/h).
2. We know |h·sin(1/h)| ≤ |h| and |h|→0 as h→0.
3. By squeeze theorem the limit is 0. So f'(0)=0, hence differentiable at 0.
Q43. Using first principles, find d/dx (1/x) for x≠0.

Formula used: f'(x)=lim_h→0[f(x+h)−f(x)]/h.
1. Compute: [1/(x+h) − 1/x]/h = [(x − (x+h))/(x(x+h))]/h = [−h]/[h·x(x+h)] = −1/[x(x+h)].
2. Let h→0 → result −1/x².
3. Answer: d/dx(1/x) = −1/x².
Q44. Let g(x)=x sin(1/x) for x≠0 and g(0)=0. Show g is continuous at 0 but not differentiable there.

Formula used: Squeeze theorem for continuity; derivative by first principle to test differentiability.
1. Continuity: |x sin(1/x)| ≤ |x| → 0 as x→0. Since g(0)=0, g is continuous at 0.
2. Differentability: compute g'(0)=lim_h→0 [h sin(1/h)]/h = lim_h→0 sin(1/h).
3. But sin(1/h) oscillates between −1 and 1 and has no limit as h→0. So g'(0) does not exist.
4. Conclusion: continuous at 0 but not differentiable at 0 (corner/oscillation).
Q45. Evaluate lim_x→0 (1 − cos x)/x² using L'Hôpital's rule.

Formula used: L'Hôpital's rule (apply when 0/0 form): lim f/g = lim f'/g' if limit exists.
1. As x→0, numerator and denominator → 0 ⇒ 0/0 form. Differentiate numerator & denominator:
2. First derivative: numerator → sin x, denominator → 2x. Limit still 0/0; apply L'Hôpital again.
3. Second derivative: numerator → cos x, denominator → 2. Evaluate at 0: cos 0 / 2 = 1/2.
4. Answer: 1/2.
Q46. Find d/dx (xˣ) for x>0 using logarithmic differentiation.

Formula used: Take logs: ln y = x ln x. Differentiate implicitly.
1. Let y = xˣ. Then ln y = x ln x.
2. Differentiating: y'/y = ln x + x·(1/x) = ln x + 1.
3. So y' = y(ln x + 1) = xˣ(ln x + 1).
4. Answer: d/dx(xˣ) = xˣ(ln x + 1).
Q47. Apply Rolle's theorem to f(x)=x³ − 3x² + 2x on [0,1] and find the point(s) c∈(0,1) with f'(c)=0.

Formula used: Rolle's theorem: if f continuous on [a,b], differentiable on (a,b), and f(a)=f(b), then ∃c with f'(c)=0.
1. Compute f(0)=0 and f(1)=1−3+2=0 ⇒ endpoints equal.
2. Compute f'(x)=3x² − 6x + 2. Set to zero: 3x² −6x +2 =0.
3. Solve: x = [6 ± √(36 − 24)]/(6) = [6 ± 2√3]/6 = 1 ± √3/3. Numeric roots ≈ 0.4226 and 1.5774.
4. Only c ≈ 0.4226 lies in (0,1). So that is the required point.
Q48. Use the Mean Value Theorem for f(x)=x² on [1,3] to find c with f'(c)=(f(3)−f(1))/(3−1).

Formula used: MVT: f'(c) = (f(b)−f(a))/(b−a).
1. f(3)−f(1) = 9 − 1 = 8. Interval length = 2 → RHS = 8/2 = 4.
2. Compute f'(x) = 2x. Set 2c = 4 ⇒ c = 2.
3. Answer: c = 2.
Q49. If f(x)=eˣ, use the derivative of inverse formula to find (f⁻¹)'(1) (i.e., derivative of ln x at x=1).

Formula used: If y=f(x) and f'(a)≠0, then (f⁻¹)'(f(a)) = 1 / f'(a).
1. Take a=0 ⇒ f(a)=e⁰=1. f'(x)=eˣ so f'(0)=1.
2. Thus (f⁻¹)'(1) = 1 / 1 = 1. Since f⁻¹ = ln, we have (ln x)'|_{x=1} = 1.
Q50. For h(x)=x² sin(1/x) (h(0)=0), we found h'(0)=0. Is h' continuous at 0? Explain.

Formula used: Compute h'(x) for x≠0 and examine limit as x→0.
1. For x≠0, use product & chain rules: h'(x)=2x sin(1/x) + x²·cos(1/x)·(−1/x²) = 2x sin(1/x) − cos(1/x).
2. As x→0, 2x sin(1/x) → 0 but −cos(1/x) oscillates between −1 and 1 and has no limit.
3. Thus lim_x→0 h'(x) does not exist, while h'(0)=0. So h' is not continuous at 0.
Q51. Evaluate lim_x→0 (eˣ − 1 − x) / x².

Formula used: Taylor expansion: eˣ = 1 + x + x²/2 + O(x³).
1. Numerator ≈ (1 + x + x²/2 + ... − 1 − x) = x²/2 + O(x³).
2. Divide by x² → limit is 1/2.
3. Answer: 1/2.
Q52. Prove that lim_x→∞ (ln x) / xᵃ = 0 for any a>0.

Formula used: L'Hôpital's rule (∞/∞ form) or growth rates: powers dominate logs.
1. As x→∞, numerator & denominator → ∞. Apply L'Hôpital: differentiate numerator 1/x and denominator a x^{a−1}.
2. Ratio becomes (1/x)/(a x^{a−1}) = 1/(a x^{a}) → 0 as x→∞.
3. Conclusion: ln x</ grows much slower; limit is 0.
Q53. Differentiate implicitly: x² + y² = 25. Then find slope at (3,4).

Formula used: Implicit differentiation: differentiate both sides w.r.t. x, treating y=y(x).
1. Differentiate: 2x + 2y·(dy/dx) = 0 ⇒ dy/dx = −x/y.
2. At (3,4): dy/dx = −3/4.
Q54. For the relation xy + eʸ = x², find dy/dx implicitly.

Formula used: Differentiate both sides; use product & chain rules.
1. Differentiate LHS: d/dx(xy) + d/dx(eʸ) = x·(dy/dx) + y·1 + eʸ·(dy/dx).
2. Differentiate RHS: 2x.
3. Collect terms with dy/dx: (x + eʸ)·(dy/dx) + y = 2x.
4. So dy/dx = (2x − y) / (x + eʸ).
Q55. Is φ(x)=|x|³ differentiable at x=0? If yes, find φ'(0).

Formula used: Check left & right derivatives; use power rule on each side.
1. For x>0: φ(x)=x³ ⇒ φ'(x)=3x², so φ'₊(0)=0.
2. For x<0 code="">: φ(x)= (−x)³ = −x³ ⇒ φ'(x)= −3x², so φ'₋(0)=0.
3. Both one-sided derivatives equal 0 ⇒ differentiable at 0 and φ'(0)=0.
Q56. If f'(x)=0 for all x in an interval [a,b], prove that f is constant on [a,b].

Formula used: Mean Value Theorem: ∃c∈(x₁,x₂) such that f'(c)=(f(x₂)−f(x₁))/(x₂−x₁).
1. Take any x₁,x₂∈[a,b]. By MVT there exists c∈(x₁,x₂) with f'(c) = (f(x₂)−f(x₁))/(x₂−x₁).
2. But f'(c)=0 by hypothesis ⇒ f(x₂)−f(x₁)=0 ⇒ f(x₂)=f(x₁).
3. Since x₁,x₂ arbitrary, f is constant on [a,b].
Q57. Find d²/dx² [x sin x] (second derivative).

Formula used: Product rule twice.
1. First derivative: d/dx(x sin x) = sin x + x cos x.
2. Differentiate again: d/dx(sin x) + d/dx(x cos x) = cos x + (cos x − x sin x).
3. Simplify: = 2 cos x − x sin x.
4. Answer: d²/dx²[x sin x] = 2 cos x − x sin x.
Q58. Evaluate lim_x→0 (tan x − x) / x³.

Formula used: Series expansion: tan x = x + x³/3 + O(x⁵).
1. Numerator ≈ (x + x³/3 + ... − x) = x³/3 + O(x⁵).
2. Divide by x³ → limit is 1/3.
3. Answer: 1/3.
Q59. Differentiate y = arctan(x²).

Formula used: d/dx[arctan u] = u'/(1+u²) and chain rule.
1. Let u = x², then du/dx = 2x.
2. So dy/dx = 2x / (1 + x⁴).
3. Answer: y' = 2x/(1 + x⁴).
Q60. Find the linear approximation (first-order Taylor polynomial) of f(x)=eˣ at a=0 and use it to approximate e^{0.1}.

Formula used: First-order Taylor: f(x) ≈ f(a) + f'(a)(x−a).
1. For f(x)=eˣ, f(0)=1 and f'(x)=eˣ so f'(0)=1.
2. Linear approx at 0: L(x)=1 + 1·x = 1 + x.
3. Approximate e^{0.1} ≈ 1 + 0.1 = 1.1 (actual ≈ 1.10517; linear approx is close for small x).

Q61–Q80 (Mixed Conceptual)

Q61. Use L'Hôpital's rule to evaluate lim_x→0 (eˣ − 1 − x) / x².

Formula used: L'Hôpital's rule for 0/0: replace f/g by f'/g' while limit remains indeterminate.
1. As x→0, numerator & denominator → 0 ⇒ 0/0 form. Differentiate numerator and denominator once: numerator' = eˣ − 1, denominator' = 2x.
2. At x→0 still 0/0, apply L'Hôpital again: differentiate numerator second time = eˣ, denominator second time = 2.
3. Limit = e⁰ / 2 = 1/2.
4. Answer: 1/2.
Q62. Determine whether f(x)=|x| is continuous and differentiable on ℝ.

Formula used: continuity: left & right limits equal f(a); differentiability: left & right derivatives equal.
1. |x| is piecewise: x for x≥0, −x for x<0 code="">. Both pieces are polynomials ⇒ continuous everywhere; at 0 both one-sided limits = 0, so continuous at 0.
2. Derivative: for x>0, f'(x)=1; for x<0 code="">, f'(x)=−1. One-sided derivatives at 0 are 1 and −1 respectively, not equal.
3. Conclusion: continuous on ℝ, but not differentiable at x=0.
Q63. Evaluate d/dx (sin⁻¹( (2x)/(1+x²) )) for |x|≠1.

Formula used: derivative of arcsin u: u' / √(1 − u²); simplify using identity.
1. Let u = 2x/(1+x²). Compute u' = [2(1+x²) − 2x·2x]/(1+x²)² = [2 + 2x² − 4x²]/(1+x²)² = [2 − 2x²]/(1+x²)² = 2(1 − x²)/(1+x²)².
2. Compute 1 − u² = 1 − (4x²/(1+x²)²) = ( (1+x²)² − 4x² )/(1+x²)² = (1 + 2x² + x⁴ − 4x²)/(1+x²)² = (1 − 2x² + x⁴)/(1+x²)² = (1 − x²)²/(1+x²)².
3. So √(1 − u²) = |1 − x²|/(1+x²). For |x|≠1 use nonzero; combine:
4. Derivative = u' / √(1−u²) = [2(1 − x²)/(1+x²)²] · [(1+x²)/|1 − x²|] = 2/(1+x²) · ( (1 − x²)/|1 − x²| ).
5. Thus d/dx = 2/(1+x²) when 1 − x²>0 (|x|<1 and="" code="">−2/(1+x²)

when |x|>1, the sign factor coming from absolute value. For principal branch usually restrict |x|<1 derivative="<code">2/(1+x²).

Q64. Find the derivative of y = x² ln x for x>0.

Formula used: Product rule: (uv)' = u'v + uv'.

Let u = x², v = ln x. Then u' = 2x, v' = 1/x.
So y' = 2x·ln x + x²·(1/x) = 2x ln x + x.
Answer: y' = 2x ln x + x.

Q65. Prove that if f is differentiable at a, then g(x)=|f(x)| need not be differentiable at a. Give an example.

Formula used: Consider f(a)=0 and derivative nonzero or sign change cases.

Example: take f(x)=x at a=0. f is differentiable with f'(0)=1.
Then g(x)=|x| is not differentiable at 0 (one-sided derivatives ±1). So differentiable f can produce non-differentiable |f| when f(a)=0 and sign changes.
Conclusion: differentiability of f does not guarantee differentiability of |f|.

Q66. Use implicit differentiation to find dy/dx for sin(x + y) = xy.

Formula used: Differentiate both sides w.r.t. x, treat y=y(x), apply chain & product rules.

Differentiate LHS: cos(x + y)·(1 + dy/dx). Differentiate RHS: y + x·(dy/dx).
So cos(x + y) + cos(x + y)·(dy/dx) = y + x·(dy/dx).
Collect dy/dx terms: dy/dx·[cos(x+y) − x] = y − cos(x+y).
Therefore dy/dx = [y − cos(x+y)] / [cos(x+y) − x] (provided denominator ≠ 0).

Q67. Find intervals where f(x)=x³ − 3x² + 2 is increasing/decreasing and locate local extrema.

Formula used: First derivative test: compute f'(x) and sign changes.

f'(x)=3x² − 6x = 3x(x − 2). Critical points at x=0 and x=2.
Test intervals:
- x<0 code="">: f'(x)>0 ⇒ increasing.
- 0: f'(x)<0 code=""> (since x positive but x−2 negative) ⇒ decreasing.
- x>2: f'(x)>0 ⇒ increasing.
So local maximum at x=0 (change + → − gives local max), local minimum at x=2 (− → + gives local min). Compute values: f(0)=2, f(2)=8 − 12 + 2 = −2.

Q68. Compute lim_x→0 ( (1 + x)^{1/x} ) and identify the limit as a known constant.

Formula used: Standard limit: lim_x→0 (1 + x)^{1/x} = e.

Take natural log: consider ln L = lim_x→0 (1/x) ln(1 + x). Use expansion ln(1+x) ~ x − x²/2 + ....
So (1/x) ln(1+x) → 1. Therefore ln L = 1, so L = e.
Answer: e.

Q69. Evaluate lim_x→0 ( (sin x)/x − cos x ) / x².

Formula used: Use Taylor expansions up to required order: sin x = x − x³/6 + ..., cos x = 1 − x²/2 + x⁴/24 + ....

Compute (sin x)/x = 1 − x²/6 + O(x⁴).
So numerator = (1 − x²/6 + ...) − (1 − x²/2 + ...)= (−x²/6 + x²/2) + O(x⁴) = x²(1/2 − 1/6) + O(x⁴) = x²(1/3) + O(x⁴).
Divide by x² → limit = 1/3.

Q70. Show that derivative of inverse function satisfies (f⁻¹)'(y) = 1 / f'(x) for y=f(x), and compute (arcsin x)' using this idea.

Formula used: Inverse function derivative formula and known derivative of sin.

If y=f(x) and f'(x)≠0, then (f⁻¹)'(y) = 1 / f'(x).
Take f(x)=sin x, then f'(x)=cos x. For inverse f⁻¹ = arcsin, at point y corresponding to x, (arcsin)'(y) = 1 / cos x.
Express cos x in terms of y: cos x = √(1 − sin²x) = √(1 − y²). Therefore d/dy arcsin y = 1/√(1 − y²), as standard.

Q71. Evaluate lim_x→0 ( (1/ x) − (1/ sin x) ).

Formula used: Expand or combine into single fraction and use series or l'Hôpital.

Combine: (1/x) − (1/sin x) = [sin x − x] / (x sin x).
As x→0, numerator ~ −x³/6, denominator ~ x·x = x² so overall ~ (−x³/6)/x² = −x/6 → 0.
Answer: 0.

Q72. Let f(x)=x² sin(1/x²) for x≠0, f(0)=0. Is f differentiable at 0?

Formula used: Check derivative definition and bound with squeeze theorem.

Difference quotient: [f(h)−0]/h = [h² sin(1/h²)]/h = h sin(1/h²).
Absolute value ≤ |h| → limit 0 as h→0. So derivative exists and equals 0.
Conclusion: f is differentiable at 0 with f'(0)=0.

Q73. Find the Taylor polynomial of degree 2 for f(x)=ln(1+x) about 0 and approximate ln1.2.

Formula used: Maclaurin expansion: ln(1+x)=x − x²/2 + x³/3 − ....

Degree-2 polynomial: P₂(x)=x − x²/2.
For x=0.2 (ln1.2): P₂(0.2)=0.2 − 0.04/2 = 0.2 − 0.02 = 0.18.
Actual ln1.2 ≈ 0.18232, approximation quite close.

Q74. For f(x)=x³, compute the third derivative f'''(x) and interpret geometrically.

Formula used: Successive differentiation: f'(x)=3x², f''(x)=6x, f'''(x)=6.

Third derivative is constant 6.
Geometric meaning: third derivative relates to change of curvature (jerk in mechanics). For cubic, constant value indicates uniform rate of change of curvature.

Q75. Show that if f has local extremum at c and f is differentiable there, then f'(c)=0.

Formula used: Fermat's theorem for stationary points.

If f has local max/min at c, then for small h, f(c+h) − f(c) has same sign or ≤0/≥0. Divide by h and take limits from both sides to get one-sided derivatives ≤0 and ≥0 respectively.
Thus both one-sided derivatives are 0 ⇒ derivative exists and equals 0.
Conclusion: f'(c)=0.

Q76. Evaluate lim_x→0 ( (1 + x)^{1/x} − e ) / x (use expansion up to x² term).

Formula used: Expansion: (1+x)^{1/x} = e · (1 − x/2 + 11x²/24 + ...) or use series for ln and exponentiate.

Let L(x)=(1+x)^{1/x}. ln L = (1/x) ln(1+x) = 1 − x/2 + x²/3 + O(x³) (expand ln).
So L = e^{1 − x/2 + x²/3 + ...} = e · e^{−x/2 + x²/3 + ...} = e [1 − x/2 + (x²/3 + x²/8) + ... ] = e [1 − x/2 + 11x²/24 + ...].
Then (L − e)/x = e[ (−1/2) + O(x) ] → limit as x→0 is −e/2.
Answer: −e/2.

Q77. Differentiate y = ln( (1+x)/(1−x) ).

Formula used: Logarithmic properties and derivative d/dx ln u = u'/u.

Simplify: y = ln(1+x) − ln(1−x).
Differentiate: y' = 1/(1+x) − (−1)/(1−x) = 1/(1+x) + 1/(1−x) = [ (1−x) + (1+x) ]/(1−x²) = 2/(1−x²).
Answer: y' = 2/(1−x²).

Q78. Evaluate lim_x→0 ( (cos x)^{1/x²} ).

Formula used: Use ln and expansion: ln cos x = −x²/2 + x⁴/12 + ....

Let L = (cos x)^{1/x²}. ln L = (1/x²) ln(cos x) ≈ (1/x²)(−x²/2 + x⁴/12 + ...) = −1/2 + x²/12 + ....
As x→0, ln L → −1/2 ⇒ L → e^{−1/2} = 1/√e.
Answer: e^{−1/2} (i.e. 1/√e).

Q79. Find d/dx (arcsin x + arccos x) for |x|<1 .="" summary="">
Formula used: d/dx arcsin x = 1/√(1−x²), d/dx arccos x = −1/√(1−x²).

Derivative = 1/√(1−x²) − 1/√(1−x²) = 0.

Consistent with identity arcsin x + arccos x = π/2 (constant), derivative 0.

Q80. Let f(x)=x⁴ − 4x³. Use the second derivative test to classify critical points.

Formula used: Critical points where f'=0; second derivative sign determines min/max: f''(c)>0 ⇒ local min, <0 code=""> ⇒ local max.

Compute f'(x)=4x³ − 12x² = 4x²(x − 3). Critical points at x=0 (double) and x=3.
Compute f''(x)=12x² − 24x = 12x(x − 2).
At x=0: f''(0)=0 → inconclusive via second derivative (use higher derivative or test). Check sign around 0: for small positive x, f' changes? Alternatively, examine f: since x=0 is double root of f', it's an inflection/flat point — check f'''(0)= −24 ≠ 0 indicates inflection with horizontal tangent.
At x=3: f''(3)=12·3·1 = 36>0 ⇒ local minimum at x=3. Value f(3)=81 − 108 = −27.

Q81–Q100 (Hard / Application Based – Continuity & Differentiability)

Q81. Test the continuity of f(x) = (sin x)/x at x = 0.

Solution: lim_x→0 (sin x)/x = 1 and f(0) is undefined. If we define f(0)=1, the function becomes continuous at 0.

Q82. Check differentiability of f(x) = |x| at x = 0.

Left derivative = –1, Right derivative = +1. Since LHD ≠ RHD, f(x) is not differentiable at x = 0.

Q83. Evaluate: lim_x→0 (e^x – 1)/x.

Using expansion: e^x ≈ 1+x, ⇒ limit = 1.

Q84. Find derivative of f(x) = √(x²+1) at x = 0 using first principle.

f(x) = √(x²+1). f'(0) = lim_h→0 (√(h²+1) – 1)/h = 0.

Q85. Check continuity of f(x) = (x²–1)/(x–1) at x=1.

f(x) simplifies to x+1 for x≠1. lim_x→1 f(x) = 2 but f(1) is undefined. If f(1)=2, function becomes continuous.

Q86. Differentiate f(x) = ln|x|.

f'(x) = 1/x, x ≠ 0.

Q87. Check differentiability of f(x) = |x–2| at x = 2.

Left derivative = –1, Right derivative = +1. Not differentiable at x=2.

Q88. If f(x) = {x² for x ≤ 1, 2x+1 for x>1}, check continuity at x=1.

LHL = 1, RHL = 3, f(1) = 1. Since LHL ≠ RHL, not continuous at x=1.

Q89. Differentiate f(x) = tan^-1(√x).

f'(x) = 1/[2√x (1+x)], x>0.

Q90. Find derivative of f(x) = x^x.

Let y=x^x, ln y = x ln x. ⇒ (1/y)(dy/dx) = ln x + 1. ⇒ f'(x)=x^x(1+ln x).

Q91. Test continuity of f(x) = (sin x)/x at x=0 using L'Hôpital rule.

lim_x→0 (sin x)/x = 1. If f(0)=1, function continuous at 0.

Q92. Differentiate f(x) = log_a(x), a>0, a≠1.

f'(x) = 1 / (x ln a).

Q93. If f(x) = {x² for x<0, –x for x≥0}, check differentiability at x=0.

Left derivative = 0, Right derivative = –1. Not differentiable at x=0.

Q94. Evaluate lim_x→0 (1–cos x)/x².

Using expansion cos x ≈ 1 – x²/2, ⇒ limit = 1/2.

Q95. Differentiate f(x) = (sin x)^{cos x}.

Let y=(sin x)^{cos x}. ln y=cos x ln(sin x). Differentiate ⇒ f'(x)=(sin x)^{cos x}[–sin x ln(sin x) + cos x·cot x].

Q96. Differentiate f(x)=√(1+sin x).

f'(x) = cos x / (2√(1+sin x)).

Q97. Find derivative of f(x) = |sin x| at x = π.

At x=π, sin π=0. Left derivative = cos x = –1, Right derivative = cos x = –1. Differentiable at x=π with f'(π)=–1.

Q98. If f(x) = max{x, x²}, check continuity at x=1.

For x<1, f(x)=x²; for x>1, f(x)=x. At x=1 both give 1. Continuous at x=1.

Q99. Differentiate f(x)=a^x where a>0.

f'(x) = a^x ln a.

Q100. Show that f(x)=x³ is differentiable at all real x.

f'(x)=3x², which exists for all real x. Hence differentiable everywhere.

Q81–Q105 (JEE Level – Continuity & Differentiability)

Q81. Check continuity of f(x) = |x|/x at x = 0.

Solution: LHL = –1, RHL = +1. Since limits don’t match, not continuous at x=0.

Q82. If f(x)=x|x|, test differentiability at x=0.

LHD = 0, RHD = 0 ⇒ differentiable at 0, f'(0)=0.

Q83. Evaluate lim_x→0 (tan x – sin x)/x³.

Using series expansion, limit = 1/2.

Q84. Check differentiability of f(x) = |x–1| + |x+1| at x=1.

LHD=0, RHD=2. Not differentiable at x=1.

Q85. Find derivative of f(x) = x^x at x = 1.

f'(x) = x^x(1+ln x). At x=1 ⇒ f'(1)=1.

Q86. Evaluate lim_x→0 (e^2x – 1 – 2x)/x².

Expansion gives limit = 2.

Q87. Check continuity of f(x) = sin(1/x) at x=0.

Limit oscillates, does not exist ⇒ not continuous at x=0.

Q88. Find derivative of f(x) = log(sin x) at x=π/4.

f'(x) = cot x. ⇒ f'(π/4)=1.

Q89. Differentiate f(x) = (tan^-1x)².

f'(x) = 2(tan^-1x)/(1+x²).

Q90. Test differentiability of f(x) = |x|·sin(1/x) at x=0.

f(0)=0, limit for derivative = 0 (squeeze theorem). Differentiable at 0.

Q91. Evaluate lim_x→0 (√(1+x) – √(1–x))/x.

Multiply by conjugate ⇒ limit = 1.

Q92. If f(x) = |x|, evaluate (f∘f)'(x) at x≠0.

f(f(x))=| |x| |=|x|. ⇒ derivative = x/|x|, x≠0.

Q93. Differentiate f(x)=ln(1+tan x).

f'(x)=sec²x/(1+tan x).

Q94. If f(x)=x³ for x<0 and f(x)=x² for x≥0, check differentiability at 0.

LHD=0, RHD=0 ⇒ differentiable at 0, f'(0)=0.

Q95. Find derivative of f(x)=sin^-1(2x/√(1+4x²)).

Simplify argument ⇒ sin^-1(tan θ). ⇒ derivative = 1/(1+4x²).

Q96. Evaluate lim_x→0 (sin 3x)/(sin 5x).

Limit = 3/5.

Q97. Differentiate f(x) = (cos x)^{sin x}.

ln y=sin x ln(cos x). ⇒ f'(x)=(cos x)^{sin x}[cos x ln(cos x) – tan x·sin x].

Q98. Check continuity of f(x)=1/(1+e^1/x) at x=0.

As x→0⁺, f(x)→0; as x→0⁻, f(x)→1. Discontinuous at 0.

Q99. Differentiate f(x)=tan^-1(√(1–x²)/x).

Simplifies to cot^-1(x/√(1–x²)). ⇒ derivative = –1/(1–x²).

Q100. Show that f(x)=x|x| is differentiable everywhere.

f(x)=x² for x≥0, f(x)=–x² for x<0. Both differentiable. At 0, derivative=0. Hence differentiable ∀x.

Q101. Evaluate lim_x→0 (tan x – x)/x³.

Expansion gives limit = 1/3.

Q102. If f(x)=x sin(1/x) for x≠0, f(0)=0, check continuity at 0.

lim_x→0 f(x)=0=f(0). Continuous at 0.

Q103. Check differentiability of f(x)=max{sin x, cos x} at x=π/4.

At π/4, sin=cos. Left slope = cos x, Right slope = –sin x. Not differentiable at π/4.

Q104. Differentiate f(x)=√(1+tan x) at x=0.

f'(x)=sec²x / (2√(1+tan x)). At x=0 ⇒ 1/2.

Q105. If f(x)=|x³|, check differentiability at 0.

LHD=0, RHD=0. Differentiable at 0 with f'(0)=0.

FAQ 1–50 (Continuity & Differentiability)

Frequently asked questions from basic to advanced — each FAQ shows the formula used and a clear step-by-step answer so students can revise fast.

FAQ 1 — What is the definition of continuity at a point x = a?

Formula / Definition: f is continuous at a if lim_x→a⁻ f(x) = lim_x→a⁺ f(x) = f(a).
1. Compute left-hand limit L₋ = lim_x→a⁻ f(x).
2. Compute right-hand limit L₊ = lim_x→a⁺ f(x).
3. If both exist and L₋ = L₊ = f(a), f is continuous at a; otherwise it is discontinuous there.
FAQ 2 — What is differentiability at a point x = a?

Formula / Definition: f is differentiable at a if the limit f'(a)=lim_h→0 [f(a+h)−f(a)]/h exists (finite).
1. Compute the difference quotient [f(a+h)−f(a)]/h.
2. Take the limit as h→0. If it exists and is finite, f'(a) exists and f is differentiable at a.
FAQ 3 — Prove: differentiability implies continuity.

Formula used: f'(a)=lim_h→0 [f(a+h)−f(a)]/h
1. If f'(a) exists, then [f(a+h)−f(a)] = h·(difference quotient).
2. As h→0, RHS → 0 because difference quotient tends to finite f'(a) and h→0.
3. Thus lim_h→0 f(a+h) = f(a), so f is continuous at a.
FAQ 4 — Does continuity imply differentiability? (Counterexample)

Answer / Example: No. Consider f(x)=|x|.
1. |x| is continuous everywhere (both one-sided limits equal function value).
2. Left derivative at 0 = −1, right derivative at 0 = 1 ⇒ derivatives unequal at 0.
3. Therefore continuous at 0 but not differentiable there → continuity does not imply differentiability.
FAQ 5 — What are removable, jump and infinite discontinuities?
1. Removable: lim_x→a f(x) exists but f(a) is different or undefined (hole). Example: f(x)=(x²−1)/(x−1) at x=1.
2. Jump: Left and right limits exist but are unequal: L₋ ≠ L₊. Example: step function.
3. Infinite: One limit is infinite (vertical asymptote). Example: 1/(x−a) at x=a.
FAQ 6 — How to test continuity for a piecewise function at the joining point?
1. Compute left-hand limit using the left piece.
2. Compute right-hand limit using the right piece.
3. Compute the function value at the joining point.
4. If all three equal, the function is continuous at the join; else not.
FAQ 7 — How to test differentiability for a piecewise function at a join?
1. First check continuity (necessary condition).
2. Compute left derivative (limit of difference quotient from left) and right derivative.
3. If both one-sided derivatives exist and equal, derivative exists; otherwise not differentiable at the join.
FAQ 8 — What is left-hand and right-hand derivative?

Formulas: f'₊(a)=lim_h→0⁺[f(a+h)−f(a)]/h, f'₋(a)=lim_h→0⁻[f(a+h)−f(a)]/h.
1. If both exist and are equal, derivative exists: f'(a)=f'₊(a)=f'₋(a).
FAQ 9 — Product, Quotient and Chain rules (state formulas)

Formulas:
- Product rule: (uv)' = u'v + uv'.
- Quotient rule: (u/v)' = (u'v − uv')/v², (v ≠ 0).
- Chain rule: d/dx[f(g(x))] = f'(g(x))·g'(x).
FAQ 10 — How to differentiate y = xˣ ? (x>0)

Formula used: Logarithmic differentiation.
1. Set y = xˣ, take ln: ln y = x ln x.
2. Differentiate: y'/y = ln x + 1.
3. So y' = xˣ(ln x + 1).
FAQ 11 — Implicit differentiation: Example x² + y² = 25. Find dy/dx.

Formula used: Differentiate both sides w.r.t x, treat y = y(x): d/dx(y) = y' .
1. Differentiate: 2x + 2y·y' = 0.
2. Solve for y': y' = −x/y.
3. At (3,4) y' = −3/4 (if needed).
FAQ 12 — Derivative of inverse function: formula and example (ln x).

Formula: If y = f(x) invertible and f'(a) ≠ 0 then (f⁻¹)'(f(a)) = 1 / f'(a).
1. Example: f(x)=eˣ, f'(x)=eˣ. For a=0: f(0)=1 so (ln)'(1)=1/f'(0)=1/1=1.
FAQ 13 — State and apply the Mean Value Theorem (MVT).

Statement: If f is continuous on [a,b] and differentiable on (a,b), then ∃ c∈(a,b) such that f'(c) = (f(b)−f(a))/(b−a).
1. Example: f(x)=x² on [1,3]: RHS=(9−1)/2=4 → f'(x)=2x → 2c=4 ⇒ c=2.
FAQ 14 — Rolle's theorem and a short example.

Statement: If f continuous on [a,b], differentiable on (a,b), and f(a)=f(b), then ∃c∈(a,b) with f'(c)=0.
1. Example: f(x)=x²−1 on [−1,1]. f(−1)=f(1)=0 ⇒ f'(x)=2x ⇒ c=0 satisfies f'(0)=0.
FAQ 15 — L'Hôpital's rule: when and how to use it (0/0 or ∞/∞ cases).

Rule: If lim f(x)=lim g(x)=0 or both ∞, and derivatives exist near a, then lim f/g = lim f'/g' if latter exists.
1. Check indeterminate form (0/0 or ∞/∞).
2. Differentiate numerator and denominator separately, then take the limit.
3. Repeat if still indeterminate.
FAQ 16 — Use Taylor / Maclaurin expansions to evaluate tricky limits (example sin x − x)/x³.
1. Use series: sin x = x − x³/6 + O(x⁵).
2. So numerator ≈ −x³/6 → divide by x³ gives −1/6.
FAQ 17 — How to find the equation of tangent to y = f(x) at x = a?

Formula: Tangent line: y − f(a) = f'(a)(x − a).
1. Compute f(a) and f'(a).
2. Plug into the tangent formula above to get the line equation.
FAQ 18 — How to find equation of the normal at x = a?

Formula: Normal slope = −1 / f'(a) (if f'(a) ≠ 0).
1. Tangent slope = f'(a). Normal slope = −1/f'(a).
2. Equation: y − f(a) = (−1/f'(a))(x − a).
FAQ 19 — What is a higher-order derivative? Example of second derivative of x³.

Notation: f'(x), f''(x), f'''(x).
1. f(x)=x³ → f'(x)=3x² → f''(x)=6x → f'''(x)=6.
2. Second derivative gives information about concavity; third derivative about rate of change of concavity.
FAQ 20 — Why is |x| not differentiable at 0? (explain corner)
1. For x>0 derivative = 1; x<0 derivative="−1.</li">
2. One-sided derivatives at 0 differ (1 ≠ −1), so derivative does not exist at 0 — this is a corner point.
FAQ 21 — What is a cusp and how does it differ from a corner?
1. Corner: One-sided derivatives finite but unequal (e.g. |x| at 0).
2. Cusp: One-sided derivatives are infinite with opposite sign (sharp pointing like x^{2/3}). Example y=x^{2/3} has vertical tangent/cusp at 0.
FAQ 22 — Use squeeze theorem to evaluate lim (sin x)/x as x→0.
1. Use inequality cos x ≤ (sin x)/x ≤ 1 (or geometric argument).
2. As x→0 both bounds → 1, hence middle → 1.
FAQ 23 — How to recognise derivative form in limits: lim_x→a [f(x) − f(a)]/(x − a).
1. This is precisely the definition of f'(a) (difference quotient) if the limit exists.
2. So recognise and compute derivative instead of doing algebraic limit if allowed.
FAQ 24 — Derivatives of inverse trig functions (list).
- d/dx[sin⁻¹ x] = 1/√(1−x²) (|x|<1 li="">
- d/dx[cos⁻¹ x] = −1/√(1−x²)
- d/dx[tan⁻¹ x] = 1/(1+x²)
FAQ 25 — Differentiate aˣ and xᵃ (a constant).
1. d/dx[aˣ] = aˣ ln a.
2. d/dx[xᵃ] = a x^{a−1} when a constant.
3. For xᵃ with variable exponent use log-diff: d/dx[xˣ] = xˣ(ln x + 1).
FAQ 26 — Chain rule example: differentiate sin(x²).

Formula: d/dx f(g(x)) = f'(g(x))·g'(x).
1. Here f(u)=sin u, g(x)=x² → f'(u)=cos u, g'(x)=2x.
2. So derivative = cos(x²)·2x = 2x cos(x²).
FAQ 27 — Parametric differentiation: dy/dx for x = cos t, y = sin t.

Formula: dy/dx = (dy/dt) / (dx/dt) (if dx/dt ≠ 0).
1. dx/dt = −sin t, dy/dt = cos t.
2. dy/dx = cos t / (−sin t) = −cot t.
FAQ 28 — Example of implicit differentiation: xy + eʸ = x². Find dy/dx.
1. Differentiate both sides w.r.t x: x·y' + y + eʸ·y' = 2x.
2. Collect y' terms: y' (x + eʸ) = 2x − y.
3. So y' = (2x − y)/(x + eʸ) (provided denominator ≠ 0).
FAQ 29 — Use derivative to find intervals of increase/decrease.
1. Find f'(x) and solve f'(x)=0 (critical pts) and where undefined.
2. Test sign of f' on intervals between critical pts: + ⇒ increasing, − ⇒ decreasing.
FAQ 30 — Second derivative test for local min/max and concavity.
1. If f'(c)=0 and f''(c)>0 ⇒ local minimum (concave up).
2. If f''(c)<0 concave="" down="" li="" local="" maximum="">
3. If f''(c)=0 test is inconclusive; use higher derivatives or first derivative test.
FAQ 31 — How does sign of f'' relate to concavity and inflection points?
1. f''(x)>0 ⇒ function concave up (cup shaped).
2. f''(x)<0 cap="" concave="" down="" li="" shaped="">
3. Point where f'' changes sign is an inflection point (if f'' exists or sign changes via f').
FAQ 32 — Linear approximation and differentials: approximate √(4.1).

Method: Use linearization at a=4: L(x)=f(a)+f'(a)(x−a).
1. f(x)=√x, a=4 ⇒ f(a)=2, f'(x)=1/(2√x) ⇒ f'(4)=1/4.
2. Δx=0.1 ⇒ approximation: 2 + (1/4)(0.1) = 2 + 0.025 = 2.025.
FAQ 33 — Error estimation with differentials (use |Δy| ≈ |f'(a)|·|Δx|).
1. Use differential dy = f'(a) dx to estimate change in f.
2. Example: if dx=0.01 and f'(a)=0.5 then estimated |Δy| ≈ 0.5·0.01 = 0.005.
FAQ 34 — Derivative of arcsin at a point using inverse derivative formula.
1. If y = arcsin x, then sin y = x. Differentiate: cos y · dy/dx = 1 ⇒ dy/dx = 1/cos y.
2. But cos y = √(1 − sin² y) = √(1 − x²). So d/dx arcsin x = 1/√(1 − x²).
FAQ 35 — Differentiate y = aˣ (a > 0) and y = xᵃ (a constant).
1. d/dx[aˣ] = aˣ ln a.
2. d/dx[xᵃ] = a x^{a−1} when a is constant (power rule).
FAQ 36 — Use L'Hôpital to show lim_x→∞ ln x / xᵃ = 0 (a>0).
1. Both numerator and denominator → ∞, apply L'Hôpital: differentiate top & bottom → (1/x)/(a x^{a−1}) = 1/(a x^{a}) → 0 as x→∞.
FAQ 37 — How to check limits at infinity; concept of horizontal asymptote.
1. Compute lim_x→∞ f(x). If finite L, horizontal asymptote y = L.
2. Use dominant term analysis for rational functions: degree numerator vs denominator.
FAQ 38 — How to remove a removable discontinuity?
1. If lim_x→a f(x)=L exists but f(a) ≠ L or undefined, redefine f(a)=L to make function continuous at a.
2. Example: f(x)=(x²−1)/(x−1) redefine f(1)=2 to remove the hole.
FAQ 39 — Is y = x^{2/3} differentiable at 0?
1. Compute derivative via first principle or check f'(x) = (2/3) x^{−1/3} for x ≠ 0 which → ∞ as x→0.
2. Derivative is infinite (vertical tangent), so derivative does not exist (not finite) at 0.
FAQ 40 — How to differentiate |sin x| ?
1. Split by sign: where sin x ≥ 0 use sin x; where sin x < 0 use −sin x.
2. Differentiate each piece: d/dx(sin x)=cos x, d/dx(−sin x)=−cos x. Check continuity at zeros of sin x to see differentiability.
FAQ 41 — Use MVT to bound function differences: |sin x − sin y| ≤ |x − y|.
1. Let f(t)=sin t; by MVT ∃ c between x and y with f(x)−f(y)=f'(c)(x−y)=cos c (x−y).
2. So |sin x − sin y| = |cos c|·|x − y| ≤ 1·|x − y|.
FAQ 42 — Differentiating piecewise functions: quick checklist.
1. Check continuity at join points.
2. Compute one-sided derivatives at joins; if equal derivative exists.
3. Else not differentiable at that point.
FAQ 43 — How to find local maxima/minima using derivative tests?
1. Find f'(x) and solve f'(x)=0 (critical pts).
2. Use first derivative test (sign change) or second derivative test (f''(c) sign) to classify each critical point.
FAQ 44 — Differentiate y = ln(sin x); domain considerations.

Formula: d/dx ln(u) = u'/u.
1. y' = (cos x) / (sin x) = cot x, but valid only where sin x ≠ 0 (domain of ln(sin x)).
2. So specify domain: intervals where sin x > 0 if principal real ln is used (or use absolute value variant).
FAQ 45 — Derivatives of arcsec and arccsc (formulas)
- d/dx[sec⁻¹ x] = 1 / (|x| √(x² − 1)) for |x|>1.
- d/dx[csc⁻¹ x] = −1 / (|x| √(x² − 1)) for |x|>1.
FAQ 46 — How to approximate derivatives numerically (difference quotient)?
1. Forward difference: f'(a) ≈ [f(a+h) − f(a)]/h.
2. Central difference (more accurate): f'(a) ≈ [f(a+h) − f(a−h)]/(2h).
3. Choose small h but beware numerical round-off error.
FAQ 47 — Why derivative fails at a discontinuity?
1. Derivative requires limit of the difference quotient to exist; if f is discontinuous at a, difference quotient cannot approach a single finite value.
2. So differentiability implies continuity; discontinuity ⇒ not differentiable.
FAQ 48 — Leibniz rule (generalized product rule) for n-th derivative of uv.

Formula: dⁿ(uv)/dxⁿ = Σ_{k=0}^{n} (n choose k) u^{(k)} v^{(n−k)}.
1. Use when differentiating product repeatedly; useful in proofs and series expansions.
FAQ 49 — Can an inverse function be differentiable where f'(x)=0?
1. No: inverse derivative formula requires f'(a) ≠ 0 to compute (f⁻¹)'(f(a)) = 1/f'(a).
2. If f'(a)=0, f is locally flat and inverse may not be differentiable (may not be one-to-one in neighbourhood).
FAQ 50 — Exam tips: How to efficiently solve continuity & differentiability problems?
1. Always check domain first (points where function not defined are candidates for discontinuity).
2. For piecewise functions, compute one-sided limits and one-sided derivatives where pieces meet.
3. Memorize standard limits and derivatives (sin x / x, eˣ, ln x, inverse trig).
4. Use Taylor expansions or L'Hôpital for indeterminate limits — pick the simplest tool.
5. When in doubt, test one-sided derivatives to decide differentiability; for proofs use formal definitions.
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