Class 12 Maths · Chapter 5
Short answer:
Is chapter me NCERT ke exercises 5.1 se 5.8 tak total 20+ questions cover kiye gaye hain — continuity check karna, differentiability check karna, chain rule, implicit differentiation, inverse trig functions ke derivatives, exponential/log functions ke derivatives, logarithmic differentiation (x^x type aur multiple product/quotient functions), parametric differentiation, second-order derivatives, aur Rolle's/Lagrange's Mean Value Theorem ka verification. Ye Class 12 Maths ka sabse zyada derivative-rules wala chapter hai — Chapter 6 (Application of Derivatives) is chapter ki foundation pe hi khada hai, isliye yahan concept clear hona zaroori hai.
Continuity ka matlab hai ki function ka graph bina uthaye, ek hi stroke me draw ho sakta ho — na koi break, na koi jump. Differentiability iske ek step aage hai: function sirf continuous hi nahi, balki us point par ek well-defined tangent bhi honi chahiye (koi sharp corner ya kink nahi). Is chapter me hum pehle continuity ki formal definition aur uski algebra (sum, difference, product, quotient ke rules) dekhenge, phir differentiability, aur uske baad derivatives ki poori machinery — chain rule, implicit differentiation, inverse trig functions, exponential/log functions, logarithmic differentiation, parametric form, aur second-order derivatives. Chapter ke end me Rolle's aur Lagrange's Mean Value Theorem hai — inka geometric matlab samajhna zaroori hai, proof nahi.
Chapter 5 Summary — 5 Minute Revision
1. Continuity — Definition
Ek function f, point x = a par continuous kehlata hai agar:
| Condition | Matlab |
|---|---|
| f(a) defined ho | Point par function ki value exist karti ho |
| limx→a f(x) exist kare | LHL = RHL (left-hand limit = right-hand limit) |
| limx→a f(x) = f(a) | Limit ki value function ki actual value ke barabar ho |
↔ Table ko side me swipe karein
Agar in teeno me se ek bhi condition fail ho jaye, function us point par discontinuous hai. Interval me continuous hone ke liye function ko interval ke har point par continuous hona chahiye.
2. Algebra of Continuous Functions
Agar f aur g dono x = a par continuous hain, to:
- f + g, f − g, f·g bhi x = a par continuous honge
- f/g bhi continuous hoga, bashart g(a) ≠ 0
- Polynomial functions har jagah continuous hote hain
- sin x, cos x har jagah continuous; tan x sirf un points par discontinuous jahan cos x = 0
3. Differentiability
Function f, x = a par differentiable hai agar Left Hand Derivative (LHD) aur Right Hand Derivative (RHD) dono exist karein aur barabar hon:
LHD = limh→0⁻ [f(a+h) − f(a)] / h RHD = limh→0⁺ [f(a+h) − f(a)] / h
Important theorem: Agar function differentiable hai to woh continuous bhi hoga — lekin ulta zaroori nahi. Jaise f(x) = |x|, x = 0 par continuous hai lekin differentiable nahi (corner point hai, LHD = −1 aur RHD = 1, dono unequal).
4. Chain Rule (Composite Functions)
Agar y = f(u) aur u = g(x), to:
dy/dx = dy/du × du/dx
Jab bhi "function ke andar function" dikhe (jaise sin(x²), e^(3x), log(sin x)), chain rule lagana padega — outer function ka derivative × inner function ka derivative.
5. Implicit Differentiation
Jab y ko explicitly x ke terms me nahi likha ja sakta (jaise x² + y² = 25, ya x^y + y^x = 1), poori equation ko x ke respect me differentiate karo, y ko function of x maan kar, aur har baar y ka derivative aane par chain rule se dy/dx multiply karo. Phir dy/dx ko algebraically isolate karo.
6. Derivatives of Inverse Trigonometric Functions
| Function | Derivative | Domain restriction |
|---|---|---|
| sin⁻¹x | 1/√(1−x²) | −1 < x < 1 |
| cos⁻¹x | −1/√(1−x²) | −1 < x < 1 |
| tan⁻¹x | 1/(1+x²) | x ∈ R |
| cot⁻¹x | −1/(1+x²) | x ∈ R |
| sec⁻¹x | 1/(|x|√(x²−1)) | |x| > 1 |
| cosec⁻¹x | −1/(|x|√(x²−1)) | |x| > 1 |
↔ Table ko side me swipe karein
Sign yaad rakhne ka trick: "co-" wale (cos⁻¹, cot⁻¹, cosec⁻¹) hamesha NEGATIVE derivative dete hain.
7. Exponential aur Logarithmic Functions
d/dx(eˣ) = eˣ d/dx(aˣ) = aˣ · ln a d/dx(ln x) = 1/x d/dx(logax) = 1/(x · ln a)
8. Logarithmic Differentiation
Ye technique tab use karte hain jab:
- Function me variable base aur variable power dono ho (jaise x^x, (sin x)^x) — direct power rule kaam nahi karega
- Function bahut saare terms ka product ya quotient ho (jaise [(x−1)(x−2)] / [(x−3)(x−4)])
Method: pehle dono side natural log lagao, log ke properties (log(ab) = log a + log b, log(aⁿ) = n log a) use karke simplify karo, phir implicit differentiation se dy/dx nikalo.
9. Parametric Differentiation
Agar x = f(t) aur y = g(t) dono ek common parameter t ke terms me diye hon, to:
dy/dx = (dy/dt) / (dx/dt), jahan dx/dt ≠ 0
10. Second Order Derivative
Pehle dy/dx nikalo, phir usi ko dobara x ke respect me differentiate karo:
d²y/dx² = d/dx(dy/dx)
Implicit ya parametric functions ke liye second derivative nikalte waqt pehle derivative me bhi dy/dx present hoga — usko substitute karna padta hai.
11. Mean Value Theorems
Rolle's Theorem: Agar f, [a,b] par continuous ho, (a,b) par differentiable ho, aur f(a) = f(b), to kam se kam ek point c ∈ (a,b) aisa exist karega jahan f'(c) = 0. Geometric matlab: curve ke endpoints ki height barabar hai to beech me kahin tangent zaroor horizontal hogi.
Lagrange's Mean Value Theorem (LMVT): Agar f, [a,b] par continuous ho aur (a,b) par differentiable ho, to kam se kam ek point c ∈ (a,b) aisa exist karega jahan:
f'(c) = [f(b) − f(a)] / (b − a)
Geometric matlab: curve ke kisi point par tangent, endpoints ko jodne wali secant line ke parallel hogi.
Poore Class 12 Maths ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q20)
Q1. Check the continuity of the function f(x) = 2x² − 1 at x = 3.
Hume check karna hai ki f(3), limx→3f(x) exist karta hai aur dono barabar hain ya nahi.
f(3) = 2(3)² − 1 = 18 − 1 = 17
limx→3 f(x) = 2(3)² − 1 = 17
Kyunki f(3) = limx→3f(x) = 17, function x = 3 par continuous hai.
Q2. Find all points of discontinuity of f, where f(x) = |x| + |x−1|.
Modulus functions ke liye critical points x = 0 aur x = 1 hain, jahan expression sign badalta hai. In dono points par continuity check karte hain.
At x = 0: LHL = lim(h→0) [|0−h| + |0−h−1|] = 0 + 1 = 1
RHL = lim(h→0) [|0+h| + |0+h−1|] = 0 + 1 = 1, aur f(0) = 0 + 1 = 1
LHL = RHL = f(0) → continuous at x = 0
At x = 1: LHL = |1−h| + |1−h−1| = 1 + 0 = 1, RHL = |1+h| + |1+h−1| = 1 + 0 = 1, f(1) = 1 + 0 = 1
Dono critical points par function continuous nikla, aur baaki sab jagah f polynomial-like (piecewise linear) hai to continuous hai. Isliye f har point par continuous hai — koi discontinuity nahi.
Q3. Check differentiability of f(x) = |x−3| at x = 3.
LHD aur RHD alag-alag nikalte hain.
LHD = lim(h→0⁻) [f(3+h) − f(3)] / h = lim(h→0⁻) [|h| − 0]/h = lim(h→0⁻) (−h)/h = −1
RHD = lim(h→0⁺) [f(3+h) − f(3)] / h = lim(h→0⁺) [h − 0]/h = 1
LHD ≠ RHD (−1 ≠ 1), isliye f, x = 3 par differentiable NAHI hai — yahan sharp corner hai, halaanki function continuous hai.
Q4. Differentiate y = sin(x²) with respect to x.
Ye composite function hai — outer function sin(u), inner function u = x². Chain rule lagate hain.
dy/dx = cos(x²) · d/dx(x²)
dy/dx = cos(x²) · 2x = 2x·cos(x²)
Q5. Differentiate y = e^(sin x) with respect to x.
Outer function e^u, inner function u = sin x.
dy/dx = e^(sin x) · d/dx(sin x)
dy/dx = e^(sin x) · cos x = cos x · e^(sin x)
Q6. Find dy/dx if x² + xy + y² = 100.
y explicitly x ke terms me nahi hai, isliye implicit differentiation karte hain — poori equation ko x ke respect me differentiate karte hain, y ko function of x maan kar.
d/dx(x²) + d/dx(xy) + d/dx(y²) = d/dx(100)
2x + [x·dy/dx + y·1] + 2y·dy/dx = 0 (product rule on xy, chain rule on y²)
2x + x·dy/dx + y + 2y·dy/dx = 0
dy/dx (x + 2y) = −(2x + y)
dy/dx = −(2x + y) / (x + 2y)
Q7. Find dy/dx if sin²y + cos xy = π.
Poori equation x ke respect me differentiate karte hain.
2 sin y · cos y · dy/dx + [−sin(xy)] · d/dx(xy) = 0
sin(2y)·dy/dx − sin(xy)·[x·dy/dx + y] = 0
sin(2y)·dy/dx − x·sin(xy)·dy/dx − y·sin(xy) = 0
dy/dx [sin(2y) − x·sin(xy)] = y·sin(xy)
dy/dx = y·sin(xy) / [sin(2y) − x·sin(xy)]
Q8. Differentiate y = sin⁻¹(2x√(1−x²)) with respect to x, where −1/√2 < x < 1/√2.
Substitution se simplify karte hain: maan lo x = sin θ, to θ = sin⁻¹x.
y = sin⁻¹(2 sin θ · cos θ) = sin⁻¹(sin 2θ) = 2θ = 2 sin⁻¹x
Ab seedha differentiate karte hain.
dy/dx = 2 · 1/√(1−x²) = 2/√(1−x²)
Q9. Differentiate y = tan⁻¹[(√(1+x²) − 1)/x] with respect to x.
Substitution x = tan θ karte hain, jahan θ = tan⁻¹x.
√(1+x²) = sec θ, isliye y = tan⁻¹[(sec θ − 1)/tan θ]
= tan⁻¹[(1−cos θ)/sin θ] = tan⁻¹[tan(θ/2)] = θ/2 = (1/2) tan⁻¹x
dy/dx = (1/2) · 1/(1+x²) = 1/[2(1+x²)]
Q10. Differentiate y = eˣ² (i.e. e raised to x²) with respect to x.
Chain rule: outer eᵘ, inner u = x².
dy/dx = e^(x²) · 2x = 2x·e^(x²)
Q11. Differentiate y = log(log x), x > 1, with respect to x.
Outer function log(u), inner function u = log x. Chain rule.
dy/dx = 1/(log x) · d/dx(log x) = 1/(log x) · 1/x
dy/dx = 1/(x · log x)
Q12. Differentiate y = x^x (x to the power x), x > 0, with respect to x using logarithmic differentiation.
Yahan base bhi x hai aur power bhi x — dono variable. Direct power rule ya exponent rule kaam nahi karega, isliye logarithmic differentiation use karte hain: pehle dono side log lagao.
y = xˣ
Taking log both sides: log y = x · log x
Differentiating both sides w.r.t. x: (1/y)·dy/dx = 1·log x + x·(1/x)
(1/y)·dy/dx = log x + 1
dy/dx = y(log x + 1) = xˣ(log x + 1)
Q13. Differentiate y = (sin x)^(cos x) with respect to x using logarithmic differentiation.
Base sin x aur power cos x dono variable hain — logarithmic differentiation zaroori hai.
y = (sin x)^(cos x)
Taking log: log y = cos x · log(sin x)
Differentiating: (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) [ cos²x/sin x − sin x·log(sin x) ]
Q14. Find dy/dx of y = √[(x−1)(x−2)] / [(x−3)(x−4)(x−5)] using logarithmic differentiation.
Ye bahut saare product/quotient terms ka function hai — logarithmic differentiation se calculation kaafi simple ho jati hai.
log y = (1/2)[log(x−1) + log(x−2) − log(x−3) − log(x−4) − log(x−5)]
Differentiating both sides: (1/y)·dy/dx = (1/2)[1/(x−1) + 1/(x−2) − 1/(x−3) − 1/(x−4) − 1/(x−5)]
dy/dx = (y/2) [1/(x−1) + 1/(x−2) − 1/(x−3) − 1/(x−4) − 1/(x−5)]
Jahan y = √[(x−1)(x−2)] / [(x−3)(x−4)(x−5)] wapas substitute kar dete hain.
Q15. If x = a(θ − sin θ) and y = a(1 − cos θ), find dy/dx.
Ye parametric form hai — x aur y dono parameter θ ke terms me diye hain. Pehle dx/dθ aur dy/dθ alag-alag nikalte hain.
dx/dθ = a(1 − cos θ)
dy/dθ = a(sin θ − 0) = a sin θ
dy/dx = (dy/dθ)/(dx/dθ) = a sin θ / [a(1 − cos θ)] = sin θ / (1 − cos θ)
Using identities: sin θ = 2 sin(θ/2)cos(θ/2), 1−cos θ = 2sin²(θ/2)
dy/dx = 2 sin(θ/2)cos(θ/2) / [2 sin²(θ/2)] = cot(θ/2)
Q16. If x = a cos³t and y = a sin³t, find dy/dx.
Parametric differentiation.
dx/dt = a · 3cos²t · (−sin t) = −3a cos²t sin t
dy/dt = a · 3sin²t · cos t = 3a sin²t cos t
dy/dx = (dy/dt)/(dx/dt) = [3a sin²t cos t] / [−3a cos²t sin t] = −(sin t/cos t) = −tan t
Q17. If y = 5cos x − 3sin x, prove that d²y/dx² + y = 0.
Pehle dy/dx, phir d²y/dx² nikalte hain.
dy/dx = −5 sin x − 3 cos x
d²y/dx² = −5 cos x − 3(−sin x) = −5 cos x + 3 sin x
d²y/dx² = −(5 cos x − 3 sin x) = −y
Isliye d²y/dx² + y = −y + y = 0. Hence proved.
Q18. If y = Aeᵐˣ + Beⁿˣ, show that d²y/dx² − (m+n)dy/dx + mny = 0.
Do baar differentiate karte hain.
dy/dx = Am·e^(mx) + Bn·e^(nx)
d²y/dx² = Am²·e^(mx) + Bn²·e^(nx)
Ab LHS me substitute karte hain:
d²y/dx² − (m+n)dy/dx + mny
= [Am² e^(mx) + Bn² e^(nx)] − (m+n)[Am e^(mx) + Bn e^(nx)] + mn[Ae^(mx) + Be^(nx)]
A e^(mx) term: m² − (m+n)m + mn = m² − m² − mn + mn = 0
B e^(nx) term: n² − (m+n)n + mn = n² − mn − n² + mn = 0
Dono terms zero ho jate hain, isliye pura expression 0 hai. Hence proved.
Q19. Verify Rolle's Theorem for f(x) = x² − 4x + 3 on the interval [1, 3].
Rolle's Theorem ke teen conditions check karte hain: continuity, differentiability, aur f(a) = f(b).
f(x) = x² − 4x + 3 ek polynomial hai, to [1,3] par continuous aur (1,3) par differentiable hai — pehli do conditions satisfied.
f(1) = 1 − 4 + 3 = 0; f(3) = 9 − 12 + 3 = 0
f(1) = f(3) = 0 — teesri condition bhi satisfied.
Rolle's Theorem ke saare conditions satisfied hain, isliye kam se kam ek c ∈ (1,3) exist karega jahan f'(c) = 0. Verify karte hain:
f'(x) = 2x − 4. Setting f'(c) = 0: 2c − 4 = 0 → c = 2
c = 2, jo (1,3) ke andar hai. Theorem verified.
Q20. Verify Lagrange's Mean Value Theorem for f(x) = x² on the interval [1, 4], and find the value of c.
Pehle conditions check karte hain, phir c nikalte hain.
f(x) = x², polynomial hone ki wajah se [1,4] par continuous aur (1,4) par differentiable hai — LMVT applicable hai.
f(1) = 1, f(4) = 16
[f(4) − f(1)] / (4 − 1) = (16 − 1)/3 = 15/3 = 5
f'(x) = 2x, to f'(c) = 5 ke liye: 2c = 5 → c = 5/2
c = 5/2 = 2.5, jo (1,4) ke andar hai. LMVT verified.
Important Equations — Ek Nazar Me
| Rule / Function | Formula |
|---|---|
| Chain rule | dy/dx = dy/du × du/dx |
| Product rule | d/dx(uv) = u'v + uv' |
| Quotient rule | d/dx(u/v) = (u'v − uv') / v² |
| d/dx(xⁿ) | n·x^(n−1) |
| d/dx(sin x) | cos x |
| d/dx(cos x) | −sin x |
| d/dx(tan x) | sec²x |
| d/dx(cot x) | −cosec²x |
| d/dx(sec x) | sec x·tan x |
| d/dx(cosec x) | −cosec x·cot x |
| d/dx(eˣ) | eˣ |
| d/dx(aˣ) | aˣ · ln a |
| d/dx(ln x) | 1/x |
| d/dx(logax) | 1/(x·ln a) |
| d/dx(sin⁻¹x) | 1/√(1−x²) |
| d/dx(cos⁻¹x) | −1/√(1−x²) |
| d/dx(tan⁻¹x) | 1/(1+x²) |
| d/dx(cot⁻¹x) | −1/(1+x²) |
| d/dx(sec⁻¹x) | 1/(|x|√(x²−1)) |
| d/dx(cosec⁻¹x) | −1/(|x|√(x²−1)) |
| Rolle's Theorem — conditions | f continuous on [a,b], differentiable on (a,b), f(a) = f(b) |
| Rolle's Theorem — conclusion | ∃ c ∈ (a,b) such that f'(c) = 0 |
| Lagrange's MVT — conditions | f continuous on [a,b], differentiable on (a,b) |
| Lagrange's MVT — conclusion | ∃ c ∈ (a,b) such that f'(c) = [f(b) − f(a)] / (b − a) |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Chain rule bhool jana. Composite function differentiate karte waqt (jaise sin(x²), e^(3x)) students sirf outer function differentiate karke reh jate hain, inner function ka derivative multiply karna bhool jate hain. Har composite function me poochho: 'kya isme function ke andar function hai?' — agar haan, chain rule lagega.
- Inverse trig derivatives me sign ki galti. cos⁻¹x, cot⁻¹x, aur cosec⁻¹x ke derivatives NEGATIVE hote hain, jabki sin⁻¹x, tan⁻¹x, sec⁻¹x ke POSITIVE. Students in dono groups ko mix kar dete hain, khaas kar cos⁻¹x ka derivative positive likh dete hain.
- Variable power par direct power rule laga dena. x^x jaisa function dekh kar students d/dx(xⁿ) = n·x^(n−1) wala rule laga dete hain, jo galat hai kyunki wahan power constant honi chahiye. Jab base aur power dono variable hon, logarithmic differentiation hi sahi tareeka hai.
- Product rule bhool jana jab dono factors x ke function hon. Implicit differentiation me xy jaisa term aata hai to students isko sirf x·(dy/dx) likh dete hain, y·1 wala term (product rule ka doosra part) chhod dete hain.
- Rolle's Theorem ki conditions check kiye bina laga dena. Kai students seedha f'(c) = 0 solve karne baith jate hain bina check kiye ki f(a) = f(b) hai ya nahi, ya function differentiable hai ya nahi. Agar conditions satisfied nahi hain to theorem apply hi nahi hota, chahe koi c mil bhi jaye.
- Continuity aur differentiability ko ek maan lena. Students soch lete hain ki agar function continuous hai to differentiable bhi hoga. Ye ulta sach hai — differentiable hona continuous hone ko imply karta hai, lekin continuous hona differentiable hone ko guarantee nahi karta (jaise |x| at x=0).
Board-Style Important Questions
- 1 mark: Is the function f(x) = |x| continuous at x = 0? Justify.
- 2 marks: Find dy/dx if y = sin(x² + 1).
- 2 marks: Check whether the function f(x) = [x] (greatest integer function) is continuous at x = 2.
- 3 marks: Find dy/dx if x^y = y^x.
- 3 marks: Differentiate (log x)^x + x^(log x) with respect to x.
- 5 marks: Verify Rolle's Theorem for f(x) = sin x − sin 2x on [0, π], and find the value of c.
Aksar Poochhe Jaane Wale Sawaal
Continuity aur differentiability me kya farak hai?
Continuity ka matlab hai function ka graph bina break ke draw ho sake. Differentiability ka matlab hai ki us point par ek well-defined tangent bhi ho, koi sharp corner nahi. Har differentiable function continuous hota hai, lekin har continuous function differentiable nahi hota — jaise |x| function x=0 par continuous hai par differentiable nahi.
Logarithmic differentiation kab use karte hain?
Jab function me variable base aur variable power dono ho (jaise x^x, (sin x)^cos x), ya jab function bahut saare terms ka lamba product ya quotient ho. Dono side log lagane se calculation simple ho jati hai kyunki log, multiplication ko addition me convert kar deta hai.
Rolle's Theorem aur Lagrange's Mean Value Theorem me kya relation hai?
Rolle's Theorem, Lagrange's MVT ka special case hai jab f(a) = f(b) ho. Us case me Lagrange's formula [f(b)−f(a)]/(b−a) zero ho jata hai, aur hume f'(c) = 0 milta hai, jo exactly Rolle's Theorem ka result hai.
Implicit differentiation aur normal differentiation me kya farak hai?
Normal differentiation tab karte hain jab y explicitly x ke terms me diya ho (jaise y = x²). Implicit differentiation tab karte hain jab x aur y ek equation me mixed hon (jaise x² + y² = 25) aur y ko x ke terms me alag se likhna mushkil ya impossible ho. Dono me chain rule ka use hota hai, bas implicit me y ko bhi function of x maan kar differentiate karte hain.
Second order derivative ka practical matlab kya hai?
Pehla derivative dy/dx function ki slope (rate of change) batata hai. Second derivative d²y/dx² batata hai ki woh slope khud kitni tezi se badal raha hai — yani curve concave up hai ya concave down. Agle chapter (Application of Derivatives) me isi se maxima-minima nikalte hain.
Parametric differentiation me seedha dy/dx kyun nahi nikal sakte?
Kyunki parametric form me y directly x ka function nahi hota — dono x aur y ek third variable (parameter, jaise t ya θ) ke through connected hote hain. Isliye pehle dy/dt aur dx/dt alag nikalte hain, phir unka ratio (dy/dt)/(dx/dt) hi dy/dx deta hai — ye chain rule ka hi ek application hai.
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