Blog Posts
Continuity and Differentiability
By
Class 12MathsChapter 5NCERTCBSE
Continuity and Differentiability – Class 12 Maths Chapter 5
📌 Quick Overview: Chapter 5 is the largest and most important chapter in Class 12 Maths. It covers continuity of functions, differentiability, all standard differentiation rules, chain rule, implicit differentiation, parametric differentiation, logarithmic differentiation, derivatives of exponential and logarithmic functions, higher order derivatives, and Rolle's and Mean Value Theorems. This chapter is tested heavily in every CBSE board exam.
Topics Covered:
- > Continuity of a Function
- > Algebra of Continuous Functions
- > Differentiability
- > Chain Rule
- > Derivatives of Inverse Trig
- > Implicit Differentiation
- > Parametric Differentiation
- > Logarithmic Differentiation
- > Exponential Functions
- > Logarithmic Functions
- > Higher Order Derivatives
- > Rolle's Theorem
- > Mean Value Theorem
- > All Standard Derivatives
1. Continuity of a Function
Definition: A function f is said to be continuous at x = c if:
lim(x→c) f(x) = f(c)
This means three conditions must hold:
(i) f(c) is defined (exists)
(ii) lim(x→c) f(x) exists
(iii) lim(x→c) f(x) = f(c)
lim(x→c) f(x) = f(c)
This means three conditions must hold:
(i) f(c) is defined (exists)
(ii) lim(x→c) f(x) exists
(iii) lim(x→c) f(x) = f(c)
⭐ Left Hand Limit and Right Hand Limit:
For continuity at x = c:
LHL = lim(x→c⁻) f(x) = lim(h→0) f(c-h)
RHL = lim(x→c⁺) f(x) = lim(h→0) f(c+h)
f is continuous at x = c if and only if: LHL = RHL = f(c)
For continuity at x = c:
LHL = lim(x→c⁻) f(x) = lim(h→0) f(c-h)
RHL = lim(x→c⁺) f(x) = lim(h→0) f(c+h)
f is continuous at x = c if and only if: LHL = RHL = f(c)
Example 1: Check continuity of f(x) = 2x + 3 at x = 1.
f(1) = 2(1) + 3 = 5
lim(x→1) (2x+3) = 2(1)+3 = 5
Since lim = f(1) = 5, f is continuous at x = 1.
f(1) = 2(1) + 3 = 5
lim(x→1) (2x+3) = 2(1)+3 = 5
Since lim = f(1) = 5, f is continuous at x = 1.
Example 2: f(x) = {x + 2, if x is less than or equal to 3 / x - 2, if x is greater than 3}. Check continuity at x = 3.
f(3) = 3 + 2 = 5
LHL = lim(x→3⁻) (x+2) = 3+2 = 5
RHL = lim(x→3⁺) (x-2) = 3-2 = 1
LHL ≠ RHL, so f is discontinuous at x = 3.
f(3) = 3 + 2 = 5
LHL = lim(x→3⁻) (x+2) = 3+2 = 5
RHL = lim(x→3⁺) (x-2) = 3-2 = 1
LHL ≠ RHL, so f is discontinuous at x = 3.
1.1 Algebra of Continuous Functions
If f and g are continuous at x = c, then:
> f + g is continuous at x = c
> f - g is continuous at x = c
> f . g is continuous at x = c
> f/g is continuous at x = c (provided g(c) ≠ 0)
> Every polynomial function is continuous everywhere on R.
> Every rational function is continuous at every point in its domain.
> f + g is continuous at x = c
> f - g is continuous at x = c
> f . g is continuous at x = c
> f/g is continuous at x = c (provided g(c) ≠ 0)
> Every polynomial function is continuous everywhere on R.
> Every rational function is continuous at every point in its domain.
⭐ Key Fact: If f is continuous at c and g is continuous at f(c), then the composite function gof is continuous at c.
This means: sin(x²), e^(cos x), log(1+x²) — all are continuous everywhere in their domains.
This means: sin(x²), e^(cos x), log(1+x²) — all are continuous everywhere in their domains.
2. Differentiability
Definition: A function f is said to be differentiable at x = c if:
f'(c) = lim(h→0) [f(c+h) - f(c)] / h
exists (is finite). This limit is called the derivative of f at c.
f'(c) = lim(h→0) [f(c+h) - f(c)] / h
exists (is finite). This limit is called the derivative of f at c.
⭐ Left and Right Derivatives:
LHD = lim(h→0⁻) [f(c+h) - f(c)] / h
RHD = lim(h→0⁺) [f(c+h) - f(c)] / h
f is differentiable at x = c if and only if LHD = RHD.
LHD = lim(h→0⁻) [f(c+h) - f(c)] / h
RHD = lim(h→0⁺) [f(c+h) - f(c)] / h
f is differentiable at x = c if and only if LHD = RHD.
Theorem: Every differentiable function is continuous. But a continuous function may NOT be differentiable.
Differentiable → Continuous (always true)
Continuous → Differentiable (NOT always true)
Differentiable → Continuous (always true)
Continuous → Differentiable (NOT always true)
Classic Example: f(x) = |x| is continuous at x = 0 but NOT differentiable at x = 0.
LHD at 0 = lim(h→0) [|0+h| - |0|]/h = lim(h→0) |h|/h = lim(-h/h) = -1
RHD at 0 = lim(h→0⁺) h/h = 1
LHD ≠ RHD → NOT differentiable at x = 0.
LHD at 0 = lim(h→0) [|0+h| - |0|]/h = lim(h→0) |h|/h = lim(-h/h) = -1
RHD at 0 = lim(h→0⁺) h/h = 1
LHD ≠ RHD → NOT differentiable at x = 0.
3. Standard Derivatives – Master Table ⭐
| Function f(x) | Derivative f'(x) |
|---|---|
| x^n | nx^(n-1) |
| sin x | cos x |
| cos x | -sin x |
| tan x | sec² x |
| cot x | -cosec² x |
| sec x | sec x tan x |
| cosec x | -cosec x cot x |
| sin⁻¹ x | 1/sqrt(1-x²) |
| cos⁻¹ x | -1/sqrt(1-x²) |
| tan⁻¹ x | 1/(1+x²) |
| cot⁻¹ x | -1/(1+x²) |
| sec⁻¹ x | 1/(|x| sqrt(x²-1)) |
| cosec⁻¹ x | -1/(|x| sqrt(x²-1)) |
| e^x | e^x |
| a^x | a^x log a |
| log x (ln x) | 1/x |
| log_a x | 1/(x log a) |
| Constant c | 0 |
4. Rules of Differentiation
BASIC RULES:
(u + v)' = u' + v' [Sum Rule]
(u - v)' = u' - v' [Difference Rule]
(cu)' = c.u' [Constant Multiple Rule]
(uv)' = u'v + uv' [Product Rule]
(u/v)' = (u'v - uv') / v² [Quotient Rule]
4.1 Chain Rule ⭐ (Most Used)
If y = f(g(x)), then:
dy/dx = f'(g(x)) . g'(x)
In other words: differentiate the outer function, keep the inner function as-is, then multiply by derivative of inner function.
dy/dx = f'(g(x)) . g'(x)
In other words: differentiate the outer function, keep the inner function as-is, then multiply by derivative of inner function.
Example 1: y = sin(x²)
dy/dx = cos(x²) . 2x = 2x cos(x²)
dy/dx = cos(x²) . 2x = 2x cos(x²)
Example 2: y = (3x + 5)⁷
dy/dx = 7(3x+5)⁶ . 3 = 21(3x+5)⁶
dy/dx = 7(3x+5)⁶ . 3 = 21(3x+5)⁶
Example 3: y = e^(sin x)
dy/dx = e^(sin x) . cos x = cos x . e^(sin x)
dy/dx = e^(sin x) . cos x = cos x . e^(sin x)
5. Implicit Differentiation
When y is not expressed explicitly as a function of x (e.g., x² + y² = 25), we differentiate both sides with respect to x, treating y as a function of x, and solve for dy/dx.
Example: Find dy/dx if x² + y² = 25.
Differentiating both sides w.r.t. x:
2x + 2y(dy/dx) = 0
2y(dy/dx) = -2x
dy/dx = -x/y
Differentiating both sides w.r.t. x:
2x + 2y(dy/dx) = 0
2y(dy/dx) = -2x
dy/dx = -x/y
Example 2: Find dy/dx if x³ + y³ = 3xy.
3x² + 3y²(dy/dx) = 3[y + x(dy/dx)]
3x² + 3y²(dy/dx) = 3y + 3x(dy/dx)
3y²(dy/dx) - 3x(dy/dx) = 3y - 3x²
dy/dx(3y² - 3x) = 3y - 3x²
dy/dx = (y - x²)/(y² - x)
3x² + 3y²(dy/dx) = 3[y + x(dy/dx)]
3x² + 3y²(dy/dx) = 3y + 3x(dy/dx)
3y²(dy/dx) - 3x(dy/dx) = 3y - 3x²
dy/dx(3y² - 3x) = 3y - 3x²
dy/dx = (y - x²)/(y² - x)
6. Parametric Differentiation
When x = f(t) and y = g(t) are both functions of a parameter t, then:
dy/dx = (dy/dt) / (dx/dt)
dy/dx = (dy/dt) / (dx/dt)
Example: x = a cos t, y = b sin t. Find dy/dx.
dx/dt = -a sin t
dy/dt = b cos t
dy/dx = (b cos t)/(-a sin t) = -b cos t / (a sin t) = -(b/a) cot t
dx/dt = -a sin t
dy/dt = b cos t
dy/dx = (b cos t)/(-a sin t) = -b cos t / (a sin t) = -(b/a) cot t
Example 2: x = at², y = 2at. Find dy/dx.
dx/dt = 2at, dy/dt = 2a
dy/dx = 2a/(2at) = 1/t
dx/dt = 2at, dy/dt = 2a
dy/dx = 2a/(2at) = 1/t
7. Logarithmic Differentiation ⭐
Used when the function has variable in both base and exponent, or when products/quotients of many functions are involved.
Method: Take log on both sides, differentiate using log properties, then solve for dy/dx.
Example 1: y = x^x. Find dy/dx.
Taking log: log y = x log x
Differentiating both sides w.r.t. x:
(1/y)(dy/dx) = log x + x.(1/x) = log x + 1
dy/dx = y(log x + 1) = x^x (1 + log x)
Taking log: log y = x log x
Differentiating both sides w.r.t. x:
(1/y)(dy/dx) = log x + x.(1/x) = log x + 1
dy/dx = y(log x + 1) = x^x (1 + log x)
Example 2: y = (sin x)^(cos x). Find dy/dx.
log y = cos x . log(sin x)
(1/y)(dy/dx) = -sin x . log(sin x) + cos x . (cos x/sin x)
(1/y)(dy/dx) = -sin x log(sin x) + cos²x / sin x
dy/dx = (sin x)^(cos x) [-sin x log(sin x) + cos²x / sin x]
log y = cos x . log(sin x)
(1/y)(dy/dx) = -sin x . log(sin x) + cos x . (cos x/sin x)
(1/y)(dy/dx) = -sin x log(sin x) + cos²x / sin x
dy/dx = (sin x)^(cos x) [-sin x log(sin x) + cos²x / sin x]
Example 3: y = sqrt[(x-1)(x-2) / (x-3)(x-4)]. Find dy/dx.
log y = (1/2)[log(x-1) + log(x-2) - log(x-3) - log(x-4)]
(1/y)(dy/dx) = (1/2)[1/(x-1) + 1/(x-2) - 1/(x-3) - 1/(x-4)]
dy/dx = y . (1/2)[1/(x-1) + 1/(x-2) - 1/(x-3) - 1/(x-4)]
log y = (1/2)[log(x-1) + log(x-2) - log(x-3) - log(x-4)]
(1/y)(dy/dx) = (1/2)[1/(x-1) + 1/(x-2) - 1/(x-3) - 1/(x-4)]
dy/dx = y . (1/2)[1/(x-1) + 1/(x-2) - 1/(x-3) - 1/(x-4)]
8. Derivatives of Inverse Trigonometric Functions using Substitution ⭐
| Expression | Substitution | Simplifies to | Derivative |
|---|---|---|---|
| sin⁻¹(2x sqrt(1-x²)) | x = sin t | 2 sin⁻¹ x | 2/sqrt(1-x²) |
| cos⁻¹(2x²-1) | x = cos t | 2 cos⁻¹ x | -2/sqrt(1-x²) |
| tan⁻¹(2x/(1-x²)) | x = tan t | 2 tan⁻¹ x | 2/(1+x²) |
| tan⁻¹((1-x)/(1+x)) | x = tan t | pi/4 - tan⁻¹ x | -1/(1+x²) |
| sin⁻¹((1-x²)/(1+x²)) | x = tan t | pi/2 - 2 tan⁻¹ x | -2/(1+x²) |
| tan⁻¹(sqrt((1-cos x)/(1+cos x))) | x = 2t | x/2 | 1/2 |
9. Higher Order Derivatives
Second Derivative: The derivative of dy/dx is called the second derivative, written as:
d²y/dx² = d/dx(dy/dx)
Similarly, d³y/dx³ is the third derivative, and so on.
d²y/dx² = d/dx(dy/dx)
Similarly, d³y/dx³ is the third derivative, and so on.
Example 1: y = sin x. Find d²y/dx².
dy/dx = cos x
d²y/dx² = d/dx(cos x) = -sin x
dy/dx = cos x
d²y/dx² = d/dx(cos x) = -sin x
Example 2: y = e^(3x). Find d²y/dx².
dy/dx = 3e^(3x)
d²y/dx² = 9e^(3x)
dy/dx = 3e^(3x)
d²y/dx² = 9e^(3x)
Example 3: If y = A sin x + B cos x, prove that d²y/dx² + y = 0.
dy/dx = A cos x - B sin x
d²y/dx² = -A sin x - B cos x = -(A sin x + B cos x) = -y
So d²y/dx² + y = -y + y = 0 ✓
dy/dx = A cos x - B sin x
d²y/dx² = -A sin x - B cos x = -(A sin x + B cos x) = -y
So d²y/dx² + y = -y + y = 0 ✓
10. Rolle's Theorem and Mean Value Theorem ⭐
10.1 Rolle's Theorem
Statement: If a function f satisfies:
(i) f is continuous on [a, b]
(ii) f is differentiable on (a, b)
(iii) f(a) = f(b)
Then there exists at least one c in (a, b) such that f'(c) = 0.
(i) f is continuous on [a, b]
(ii) f is differentiable on (a, b)
(iii) f(a) = f(b)
Then there exists at least one c in (a, b) such that f'(c) = 0.
Example: Verify Rolle's Theorem for f(x) = x² - 4x + 3 on [1, 3].
f(1) = 1 - 4 + 3 = 0, f(3) = 9 - 12 + 3 = 0 → f(1) = f(3) ✓
f is polynomial → continuous and differentiable everywhere ✓
f'(x) = 2x - 4 = 0 → x = 2 ∈ (1, 3) ✓
Rolle's Theorem is verified with c = 2.
f(1) = 1 - 4 + 3 = 0, f(3) = 9 - 12 + 3 = 0 → f(1) = f(3) ✓
f is polynomial → continuous and differentiable everywhere ✓
f'(x) = 2x - 4 = 0 → x = 2 ∈ (1, 3) ✓
Rolle's Theorem is verified with c = 2.
10.2 Mean Value Theorem (Lagrange's MVT)
Statement: If a function f satisfies:
(i) f is continuous on [a, b]
(ii) f is differentiable on (a, b)
Then there exists at least one c in (a, b) such that:
f'(c) = [f(b) - f(a)] / (b - a)
(i) f is continuous on [a, b]
(ii) f is differentiable on (a, b)
Then there exists at least one c in (a, b) such that:
f'(c) = [f(b) - f(a)] / (b - a)
Example: Verify MVT for f(x) = x² on [2, 4].
f(2) = 4, f(4) = 16
[f(b) - f(a)]/(b-a) = (16 - 4)/(4 - 2) = 12/2 = 6
f'(x) = 2x → f'(c) = 2c = 6 → c = 3 ∈ (2, 4) ✓
MVT verified with c = 3.
f(2) = 4, f(4) = 16
[f(b) - f(a)]/(b-a) = (16 - 4)/(4 - 2) = 12/2 = 6
f'(x) = 2x → f'(c) = 2c = 6 → c = 3 ∈ (2, 4) ✓
MVT verified with c = 3.
11. Important Board Exam Questions
Q1. Find dy/dx if y = sin⁻¹[(3x - 4x³)].
Put x = sin t, so t = sin⁻¹ x
3x - 4x³ = 3 sin t - 4 sin³t = sin 3t
y = sin⁻¹(sin 3t) = 3t = 3 sin⁻¹ x
dy/dx = 3/sqrt(1 - x²)
3x - 4x³ = 3 sin t - 4 sin³t = sin 3t
y = sin⁻¹(sin 3t) = 3t = 3 sin⁻¹ x
dy/dx = 3/sqrt(1 - x²)
Q2. If y = (tan⁻¹ x)², show that (x² + 1)² y₂ + 2x(x² + 1) y₁ = 2.
y = (tan⁻¹ x)²
y₁ = dy/dx = 2 tan⁻¹x . (1/(1+x²))
y₁(1+x²) = 2 tan⁻¹x
Differentiating again: y₂(1+x²) + y₁(2x) = 2/(1+x²)
Multiply by (1+x²): (1+x²)²y₂ + 2x(1+x²)y₁ = 2 ✓
y₁ = dy/dx = 2 tan⁻¹x . (1/(1+x²))
y₁(1+x²) = 2 tan⁻¹x
Differentiating again: y₂(1+x²) + y₁(2x) = 2/(1+x²)
Multiply by (1+x²): (1+x²)²y₂ + 2x(1+x²)y₁ = 2 ✓
Q3. Differentiate x^(sin x) + (sin x)^(cos x) with respect to x.
Let u = x^(sin x) and v = (sin x)^(cos x), so y = u + v.
For u: log u = sin x . log x
(1/u)(du/dx) = cos x . log x + sin x/x
du/dx = x^(sin x)[cos x log x + sin x/x]
For v: log v = cos x . log(sin x)
(1/v)(dv/dx) = -sin x . log(sin x) + cos x . (cos x/sin x)
dv/dx = (sin x)^(cos x)[-sin x log(sin x) + cos²x/sin x]
dy/dx = du/dx + dv/dx (add both parts above)
For u: log u = sin x . log x
(1/u)(du/dx) = cos x . log x + sin x/x
du/dx = x^(sin x)[cos x log x + sin x/x]
For v: log v = cos x . log(sin x)
(1/v)(dv/dx) = -sin x . log(sin x) + cos x . (cos x/sin x)
dv/dx = (sin x)^(cos x)[-sin x log(sin x) + cos²x/sin x]
dy/dx = du/dx + dv/dx (add both parts above)
Q4. If x = a(cos t + t sin t) and y = a(sin t - t cos t), find d²y/dx².
dx/dt = a(-sin t + sin t + t cos t) = at cos t
dy/dt = a(cos t - cos t + t sin t) = at sin t
dy/dx = (at sin t)/(at cos t) = tan t
d²y/dx² = d/dx(tan t) = (d/dt(tan t))/(dx/dt) = sec²t / (at cos t) = sec³t / at
dy/dt = a(cos t - cos t + t sin t) = at sin t
dy/dx = (at sin t)/(at cos t) = tan t
d²y/dx² = d/dx(tan t) = (d/dt(tan t))/(dx/dt) = sec²t / (at cos t) = sec³t / at
Q5. Find the value of k so that f(x) = {kx + 1, x is less than or equal to 5 / 3x - 5, x is greater than 5} is continuous at x = 5.
For continuity at x = 5: LHL = RHL = f(5)
f(5) = k(5) + 1 = 5k + 1
RHL = lim(x→5⁺)(3x - 5) = 15 - 5 = 10
For continuity: 5k + 1 = 10 → 5k = 9 → k = 9/5
f(5) = k(5) + 1 = 5k + 1
RHL = lim(x→5⁺)(3x - 5) = 15 - 5 = 10
For continuity: 5k + 1 = 10 → 5k = 9 → k = 9/5
12. Key Formulas at a Glance
CONTINUITY: LHL = RHL = f(c)
DIFFERENTIABILITY: LHD = RHD
CHAIN RULE: d/dx[f(g(x))] = f'(g(x)).g'(x)
PRODUCT RULE: (uv)' = u'v + uv'
QUOTIENT RULE: (u/v)' = (u'v - uv') / v²
PARAMETRIC: dy/dx = (dy/dt) / (dx/dt)
LOGARITHMIC: Take log both sides → differentiate
IMPORTANT:
d/dx(e^x) = e^x
d/dx(a^x) = a^x log a
d/dx(log x) = 1/x
d/dx(sin⁻¹x) = 1/sqrt(1-x²)
d/dx(tan⁻¹x) = 1/(1+x²)
ROLLE'S THEOREM: f(a)=f(b) → f'(c) = 0 for some c in (a,b)
MVT: f'(c) = [f(b)-f(a)] / (b-a) for some c in (a,b)
13. MCQ Practice (1 Mark)
1. The function f(x) = |x| is:
(a) Continuous and differentiable at x = 0
(b) Continuous but not differentiable at x = 0
(c) Neither continuous nor differentiable at x = 0
(d) Differentiable but not continuous at x = 0
Answer: (b) Continuous but not differentiable at x = 0
2. If y = log(sin x), then dy/dx =
(a) cot x (b) tan x (c) -cot x (d) sec x
Answer: (a) cot x — because (1/sin x)(cos x) = cos x/sin x = cot x
3. If y = e^x . sin x, then dy/dx =
(a) e^x sin x (b) e^x cos x (c) e^x(sin x + cos x) (d) e^x(sin x - cos x)
Answer: (c) e^x(sin x + cos x)
4. The derivative of sin⁻¹(2x sqrt(1-x²)) with respect to x is:
(a) 2/sqrt(1-x²) (b) 1/sqrt(1-x²) (c) -2/sqrt(1-x²) (d) 2/(1+x²)
Answer: (a) 2/sqrt(1-x²)
5. If y = x^x, then dy/dx =
(a) x^x (b) x^x . log x (c) x^x(1 + log x) (d) x^(x-1)
Answer: (c) x^x(1 + log x)
14. Exam Tips
- For piecewise functions, ALWAYS check LHL, RHL, and f(c) separately — do not assume continuity.
- For inverse trig differentiation questions, look for substitution patterns (x = sin t, x = tan t, x = cos t) — these convert complex expressions to simple ones.
- For x^x or (f(x))^(g(x)) type functions, always use logarithmic differentiation — there is no shortcut.
- In parametric differentiation, find dy/dt and dx/dt separately, then divide — do NOT differentiate y with respect to x directly.
- For second derivative questions involving Rolle's/MVT: verify ALL three/two conditions explicitly — partial marks are given for each condition.
- Product rule mistake to avoid: (uv)' is NOT u'v' — it is u'v + uv'.
- d/dx(e^x) = e^x (not xe^(x-1)) — the power rule does NOT apply to exponential functions.
✅ Summary: Chapter 5 is the backbone of Class 12 Maths. Almost 20-25 marks in CBSE board exams come from topics in this chapter (continuity, differentiation, and applications in Chapter 6). Master the chain rule, logarithmic differentiation, inverse trig substitution tricks, and Rolle's/MVT verification. Practice daily — this chapter rewards consistent practice more than memorisation.
Study Got – Making Studies Simple | studygot.in
Post a Comment