Simple Learning HubCLASS 9–12 · JEE · NEET
📝 Board Notes: Pass to Topperबोर्ड नोट्स: पास से टॉपर तकBoard Notes: Pass se Topper tak

Continuity and Differentiabilityसांतत्य तथा अवकलनीयताContinuity and Differentiability

Class 12 Maths · Chapter 5 · UP Boardकक्षा 12 गणित · अध्याय 5 · यूपी बोर्डClass 12 Maths · Chapter 5 · UP Board

Built by analysing many years of real UP Board papers, and updated after every board exam. Beginner or topper, it is for everyone.कई वर्षों के असली यूपी बोर्ड पेपरों के विश्लेषण से बना, हर बोर्ड परीक्षा के बाद अपडेट। शुरुआत करने वाले हों या टॉपर, सबके लिए।Kai saalon ke asli UP Board papers ko analyse karke bana, har board exam ke baad update. Beginner ho ya topper, sabke liye.

📊 Based on 7 years (2019–2026) of UP Board papers, current syllabus: 38 sets readवर्तमान सिलेबस के 7 वर्षों (2019–2026) के यूपी बोर्ड पेपरों पर आधारित: 38 सेट पढ़े गएCurrent syllabus ke 7 saal (2019–2026) ke UP Board papers par based: 38 sets padhe (2019, 2020, 2022, 2023, 2024, 2025, 2026)

🔥 Most asked in this chapterइस अध्याय में सबसे ज़्यादा पूछे गएIs chapter mein sabse zyada aaye

Ranked by how many years each topic came in the paper. Tap a topic to jump to its notes.हर टॉपिक कितने वर्ष पेपर में आया, उसी क्रम में। टॉपिक पर टैप करें, नोट्स पर पहुँचें।Har topic kitne saal paper mein aaya, usi order mein. Topic par tap kijiye, notes par pahunchiye.

1Continuity at a point / find kकिसी बिंदु पर सांतत्य / k ज्ञात करनाPoint par continuity / k nikalna🔥 Asked in 7 of 7 years🔥 7 में से 7 वर्ष पूछा गया🔥 Aaya hai 7/7 saal🔴 Very importantबहुत महत्वपूर्णBahut important
Came every yearहर वर्ष आयाHar saal aaya: 2019202020222023202420252026
Marks: 1, 2, 4, 5अंक: 1, 2, 4, 5Marks: 1, 2, 4, 52Second-order derivative: "show that" relationsद्वितीय कोटि अवकलज: "दिखाइए कि" संबंधSecond order derivative: "dikhaiye ki" relations🔥 Asked in 7 of 7 years🔥 7 में से 7 वर्ष पूछा गया🔥 Aaya hai 7/7 saal🔴 Very importantबहुत महत्वपूर्णBahut important
Came every yearहर वर्ष आयाHar saal aaya: 2019202020222023202420252026
Marks: 1, 2, 3, 4, 5, 8अंक: 1, 2, 3, 4, 5, 8Marks: 1, 2, 3, 4, 5, 83Logarithmic differentiation (x^x, x^(sin x) type)लघुगणकीय अवकलन (x^x, x^(sin x) प्रकार)Logarithmic differentiation (x^x, x^(sin x) type)🔥 Asked in 7 of 7 years🔥 7 में से 7 वर्ष पूछा गया🔥 Aaya hai 7/7 saal🔴 Very importantबहुत महत्वपूर्णBahut important
Came every yearहर वर्ष आयाHar saal aaya: 2019202020222023202420252026
Marks: 2, 3, 4, 5, 8अंक: 2, 3, 4, 5, 8Marks: 2, 3, 4, 5, 84Implicit functionsअस्पष्ट फलनImplicit functions🔥 Asked in 5 of 7 years🔥 7 में से 5 वर्ष पूछा गया🔥 Aaya hai 5/7 saal🔴 Very importantबहुत महत्वपूर्णBahut important
Came / not askedआया / नहीं आयाAaya / nahi aaya: 2019202020222023202420252026
Marks: 1, 4, 5अंक: 1, 4, 5Marks: 1, 4, 55Differentiability at a pointकिसी बिंदु पर अवकलनीयताPoint par differentiability🔥 Asked in 5 of 7 years🔥 7 में से 5 वर्ष पूछा गया🔥 Aaya hai 5/7 saal🔴 Very importantबहुत महत्वपूर्णBahut important
Came / not askedआया / नहीं आयाAaya / nahi aaya: 2019202020222023202420252026
Marks: 2, 5, 8अंक: 2, 5, 8Marks: 2, 5, 86Derivatives of inverse trig functions (with substitution)प्रतिलोम त्रिकोणमितीय फलनों के अवकलज (प्रतिस्थापन सहित)Inverse trig functions ke derivatives (substitution ke saath)🔥 Asked in 5 of 7 years🔥 7 में से 5 वर्ष पूछा गया🔥 Aaya hai 5/7 saal🔴 Very importantबहुत महत्वपूर्णBahut important
Came / not askedआया / नहीं आयाAaya / nahi aaya: 2019202020222023202420252026
Marks: 1, 2, 5अंक: 1, 2, 5Marks: 1, 2, 5
Show all 11 topicsसभी 11 टॉपिक देखेंSaare 11 topics dekhiye
7Continuity of standard functionsमानक फलनों का सांतत्यStandard functions ki continuity🔥 Asked in 5 of 7 years🔥 7 में से 5 वर्ष पूछा गया🔥 Aaya hai 5/7 saal🔴 Very importantबहुत महत्वपूर्णBahut important
Came / not askedआया / नहीं आयाAaya / nahi aaya: 2019202020222023202420252026
Marks: 1, 2, 5अंक: 1, 2, 5Marks: 1, 2, 58Parametric differentiationप्राचलिक अवकलनParametric differentiationAsked in 4 of 7 years7 में से 4 वर्ष पूछा गयाAaya hai 4/7 saal
Came / not askedआया / नहीं आयाAaya / nahi aaya: 2019202020222023202420252026
Marks: 1, 2, 4, 5अंक: 1, 2, 4, 5Marks: 1, 2, 4, 59Chain ruleशृंखला नियमChain ruleAsked in 3 of 7 years7 में से 3 वर्ष पूछा गयाAaya hai 3/7 saal
Came / not askedआया / नहीं आयाAaya / nahi aaya: 2019202020222023202420252026
Marks: 1, 2, 4अंक: 1, 2, 4Marks: 1, 2, 410Derivatives of exponential and log functionsचरघातांकी तथा लघुगणकीय फलनों के अवकलजExponential aur log functions ke derivativesAsked in 3 of 7 years7 में से 3 वर्ष पूछा गयाAaya hai 3/7 saal
Came / not askedआया / नहीं आयाAaya / nahi aaya: 2019202020222023202420252026
Marks: 2, 5अंक: 2, 5Marks: 2, 5
11Rolle's theorem and Mean Value TheoremAsked in 2 of 7 years7 में से 2 वर्ष पूछा गयाAaya hai 2/7 saal
Came / not askedआया / नहीं आयाAaya / nahi aaya: 2019202020222023202420252026
Marks: 2, 5अंक: 2, 5Marks: 2, 5
🟢 PASSपासPASS

Haven't studied anything? Read just this much and you will pass.कुछ नहीं पढ़ा? बस इतना पढ़ लीजिए, पास हो जाएँगे।Kuch nahi padha? Bas itna padh lijiye, pass ho jayenge.

👉 Read this firstसबसे पहले यह पढ़ेंSabse pehle yeh padhiye

This is the biggest-scoring chapter of Class 12 calculus. Three types repeat almost every year: (1) continuity at a point / find k (LHL = RHL = f(a)); (2) second derivative 'show that' questions (y = sin⁻¹x, y = e^(tan⁻¹x), y = Ae^(mx) + Be^(nx)); (3) logarithmic differentiation of x^(sin x) + (cos x)^(tan x) type. Learn the derivative table first, then practise these three types until each takes you under 10 minutes.यह कक्षा 12 कलन का सबसे अधिक अंक देने वाला अध्याय है। तीन प्रकार लगभग हर वर्ष दोहराए जाते हैं: (1) किसी बिंदु पर सांतत्य / k ज्ञात करना (बायाँ सीमा = दायाँ सीमा = f(a)); (2) द्वितीय अवकलज वाले 'दिखाइए कि' प्रश्न (y = sin⁻¹x, y = e^(tan⁻¹x), y = Ae^(mx) + Be^(nx)); (3) x^(sin x) + (cos x)^(tan x) प्रकार का लघुगणकीय अवकलन। पहले अवकलज तालिका याद कीजिए, फिर इन तीनों का अभ्यास तब तक कीजिए जब तक प्रत्येक 10 मिनट से कम में न हो जाए।Yeh Class 12 calculus ka sabse zyada marks dene wala chapter hai. Teen types lagbhag har saal repeat hote hain: (1) kisi point par continuity / k nikalna (LHL = RHL = f(a)); (2) second derivative wale 'dikhaiye ki' sawaal (y = sin⁻¹x, y = e^(tan⁻¹x), y = Ae^(mx) + Be^(nx)); (3) x^(sin x) + (cos x)^(tan x) type ka logarithmic differentiation. Pehle derivative table yaad kijiye, phir in teeno ki practice tab tak kijiye jab tak har ek 10 minute se kam mein na ho jaaye.

🏅 Pass-guarantee minimumपास होने के लिए न्यूनतमPass hone ke liye minimum

  • Learn the derivative table (xⁿ, sin, cos, tan, eˣ, log x, sin⁻¹, tan⁻¹) by heart; MCQs and every question here use it.
  • Learn the LHL = RHL = f(a) method and solve 3 'find k' questions.
  • Learn the three second-derivative results (sin⁻¹x, e^(tan⁻¹x), Ae^(mx) + Be^(nx)) with the cross-multiply trick.
  • Practise 3 logarithmic differentiation sums like x^(cos x) + (sin x)^x (u + v separately).
  • Learn dy/dx = (dy/dt)/(dx/dt) for parametric and one implicit example.
  • अवकलज तालिका (xⁿ, sin, cos, tan, eˣ, log x, sin⁻¹, tan⁻¹) कंठस्थ कीजिए; MCQ तथा यहाँ के सभी प्रश्न इसी पर हैं।
  • बायाँ सीमा = दायाँ सीमा = f(a) विधि सीखिए और 3 'k ज्ञात कीजिए' प्रश्न हल कीजिए।
  • तीन द्वितीय अवकलज परिणाम (sin⁻¹x, e^(tan⁻¹x), Ae^(mx) + Be^(nx)) वज्र-गुणन युक्ति के साथ सीखिए।
  • x^(cos x) + (sin x)^x जैसे 3 लघुगणकीय अवकलन प्रश्नों का अभ्यास कीजिए (u + v अलग)।
  • प्राचलिक के लिए dy/dx = (dy/dt)/(dx/dt) तथा एक अस्पष्ट फलन उदाहरण सीखिए।
  • Derivative table (xⁿ, sin, cos, tan, eˣ, log x, sin⁻¹, tan⁻¹) zubaani yaad kijiye; MCQ aur yahan ke saare sawaal isi par hain.
  • LHL = RHL = f(a) method seekhiye aur 3 'k nikaliye' sawaal kijiye.
  • Teen second derivative results (sin⁻¹x, e^(tan⁻¹x), Ae^(mx) + Be^(nx)) cross-multiply trick ke saath seekhiye.
  • x^(cos x) + (sin x)^x jaise 3 logarithmic differentiation sawaalon ki practice kijiye (u + v alag).
  • Parametric ke liye dy/dx = (dy/dt)/(dx/dt) aur ek implicit example seekhiye.

🎯 1-mark facts that keep comingबार-बार आने वाले 1 अंक के तथ्यBaar baar aane wale 1 mark facts

  • d/dx sin(x² + 5) = 2x cos(x² + 5).
  • d/dx log(log x) = 1/(x log x).
  • |x| is continuous everywhere but not differentiable at x = 0.
  • Every differentiable function is continuous.
  • x = at², y = 2at ⇒ dy/dx = 1/t.
  • d/dx tan⁻¹x = 1/(1 + x²); d/dx sin⁻¹x = 1/√(1 − x²).
  • d/dx aˣ = aˣ log a.
  • d/dx tan⁻¹[sin x/(1 + cos x)] = 1/2.
  • d/dx sin(x² + 5) = 2x cos(x² + 5)।
  • d/dx log(log x) = 1/(x log x)।
  • |x| सर्वत्र संतत है परन्तु x = 0 पर अवकलनीय नहीं।
  • प्रत्येक अवकलनीय फलन संतत है।
  • x = at², y = 2at ⇒ dy/dx = 1/t।
  • d/dx tan⁻¹x = 1/(1 + x²); d/dx sin⁻¹x = 1/√(1 − x²)।
  • d/dx aˣ = aˣ log a।
  • d/dx tan⁻¹[sin x/(1 + cos x)] = 1/2।
  • d/dx sin(x² + 5) = 2x cos(x² + 5).
  • d/dx log(log x) = 1/(x log x).
  • |x| har jagah continuous hai lekin x = 0 par differentiable nahi.
  • Har differentiable function continuous hai.
  • x = at², y = 2at ⇒ dy/dx = 1/t.
  • d/dx tan⁻¹x = 1/(1 + x²); d/dx sin⁻¹x = 1/√(1 − x²).
  • d/dx aˣ = aˣ log a.
  • d/dx tan⁻¹[sin x/(1 + cos x)] = 1/2.
🟡 SCOREस्कोरSCORE

The questions that come every year, with model answers. Most asked first.हर साल आने वाले प्रश्न, आदर्श उत्तरों के साथ। सबसे ज़्यादा पूछे गए पहले।Har saal aane wale sawaal, model answers ke saath. Sabse zyada aaye pehle.

Continuity at a point / find kकिसी बिंदु पर सांतत्य / k ज्ञात करनाPoint par continuity / k nikalna

🔥 Asked in 7 of 7 years🔥 7 में से 7 वर्ष पूछा गया🔥 Aaya hai 7/7 saal🔴 Very importantबहुत महत्वपूर्णBahut important
Came every yearहर वर्ष आयाHar saal aaya: 2019202020222023202420252026
  • f is continuous at x = a if LHL = RHL = f(a), i.e. lim(x→a⁻) f(x) = lim(x→a⁺) f(x) = f(a).
  • Method for a piecewise function: (1) write f(a) from the rule that includes '='; (2) LHL: put x = a − h, h → 0, in the left rule; (3) RHL: put x = a + h in the right rule; (4) compare.
  • 'Find k': set the two sides equal (LHL = RHL = f(a)) and solve for k.
  • Two unknowns a, b with one meeting point give one equation; if f(a) is also given separately you get two equations.
  • Removable type: f(x) = (x² − 9)/(x − 3), x ≠ 3, f(3) = k ⇒ limit = 6, so k = 6.
  • If no value of k can make LHL = RHL (e.g. λ(x² − 2x) for x ≤ 0 and 4x + 1 for x > 0 at x = 0: LHL = 0, RHL = 1), write 'f is discontinuous at x = 0 for every λ'.
  • Squeeze idea: x² sin(1/x) → 0 as x → 0 because |x² sin(1/x)| ≤ x².
  • f, x = a पर संतत है यदि बायाँ सीमा = दायाँ सीमा = f(a), अर्थात lim(x→a⁻) f(x) = lim(x→a⁺) f(x) = f(a)।
  • खंडशः फलन के लिए विधि: (1) '=' वाले नियम से f(a) लिखिए; (2) बायाँ सीमा: बाएँ नियम में x = a − h, h → 0 रखिए; (3) दायाँ सीमा: दाएँ नियम में x = a + h रखिए; (4) तुलना कीजिए।
  • 'k ज्ञात कीजिए': दोनों पक्ष बराबर रखिए (बायाँ सीमा = दायाँ सीमा = f(a)) और k हल कीजिए।
  • दो अज्ञात a, b तथा एक मिलन बिंदु से एक समीकरण मिलता है; यदि f(a) अलग से दिया हो तो दो समीकरण।
  • हटाने योग्य प्रकार: f(x) = (x² − 9)/(x − 3), x ≠ 3, f(3) = k ⇒ सीमा = 6, अतः k = 6।
  • यदि k का कोई मान बायाँ सीमा = दायाँ सीमा न कर सके (जैसे x = 0 पर x ≤ 0 के लिए λ(x² − 2x) तथा x > 0 के लिए 4x + 1: बायाँ सीमा = 0, दायाँ सीमा = 1), तो लिखिए 'λ के किसी भी मान के लिए f, x = 0 पर असंतत है'।
  • सैंडविच विचार: x → 0 पर x² sin(1/x) → 0 क्योंकि |x² sin(1/x)| ≤ x²।
  • f, x = a par continuous hai agar LHL = RHL = f(a), yaani lim(x→a⁻) f(x) = lim(x→a⁺) f(x) = f(a).
  • Piecewise function ke liye method: (1) '=' wale rule se f(a) likhiye; (2) LHL: left rule mein x = a − h, h → 0 rakhiye; (3) RHL: right rule mein x = a + h rakhiye; (4) compare kijiye.
  • 'k nikaliye': dono sides barabar rakhiye (LHL = RHL = f(a)) aur k solve kijiye.
  • Do unknowns a, b aur ek meeting point se ek equation milti hai; agar f(a) alag se diya ho to do equations.
  • Removable type: f(x) = (x² − 9)/(x − 3), x ≠ 3, f(3) = k ⇒ limit = 6, isliye k = 6.
  • Agar k ki koi value LHL = RHL na kar sake (jaise x = 0 par x ≤ 0 ke liye λ(x² − 2x) aur x > 0 ke liye 4x + 1: LHL = 0, RHL = 1), to likhiye 'λ ki kisi bhi value ke liye f, x = 0 par discontinuous hai'.
  • Squeeze idea: x → 0 par x² sin(1/x) → 0 kyunki |x² sin(1/x)| ≤ x².
✏️ Formulas to rememberयाद रखने वाले सूत्रYaad rakhne wale formulas
Continuous at a ⇔ lim(x→a⁻) f = lim(x→a⁺) f = f(a)
LHL = lim(h→0) f(a − h), RHL = lim(h→0) f(a + h)
Repeated questionबार-बार पूछा गया प्रश्नBaar baar aaya sawaal · 5 marks5 अंक5 marks

Worked example: Find k if f(x) = kx + 1 for x ≤ 5 and f(x) = 3x − 5 for x > 5 is continuous at x = 5.हल किया उदाहरण: k ज्ञात कीजिए यदि f(x) = kx + 1, x ≤ 5 तथा f(x) = 3x − 5, x > 5, x = 5 पर संतत है।Solved example: k nikaliye agar f(x) = kx + 1, x ≤ 5 aur f(x) = 3x − 5, x > 5, x = 5 par continuous hai.

✅ Model answerआदर्श उत्तरModel answer
  • f(5) = 5k + 1.
  • LHL = lim(h→0) [k(5 − h) + 1] = 5k + 1.
  • RHL = lim(h→0) [3(5 + h) − 5] = 15 − 5 = 10.
  • For continuity: 5k + 1 = 10 ⇒ 5k = 9 ⇒ k = 9/5.
  • f(5) = 5k + 1।
  • बायाँ सीमा = lim(h→0) [k(5 − h) + 1] = 5k + 1।
  • दायाँ सीमा = lim(h→0) [3(5 + h) − 5] = 15 − 5 = 10।
  • सांतत्य के लिए: 5k + 1 = 10 ⇒ 5k = 9 ⇒ k = 9/5।
  • f(5) = 5k + 1.
  • LHL = lim(h→0) [k(5 − h) + 1] = 5k + 1.
  • RHL = lim(h→0) [3(5 + h) − 5] = 15 − 5 = 10.
  • Continuity ke liye: 5k + 1 = 10 ⇒ 5k = 9 ⇒ k = 9/5.
Repeated questionबार-बार पूछा गया प्रश्नBaar baar aaya sawaal · 5 marks5 अंक5 marks

Test the continuity of f(x) = −2 (x ≤ −1), 2x (−1 < x ≤ 1), 2 (x > 1) at x = −1 and x = 1.f(x) = −2 (x ≤ −1), 2x (−1 < x ≤ 1), 2 (x > 1) के सांतत्य की x = −1 तथा x = 1 पर जाँच कीजिए।f(x) = −2 (x ≤ −1), 2x (−1 < x ≤ 1), 2 (x > 1) ki continuity x = −1 aur x = 1 par check kijiye.

✅ Model answerआदर्श उत्तरModel answer
  • At x = −1: f(−1) = −2; LHL = −2; RHL = lim 2(−1 + h) = −2. All equal ⇒ continuous at x = −1.
  • At x = 1: f(1) = 2(1) = 2; LHL = lim 2(1 − h) = 2; RHL = 2. All equal ⇒ continuous at x = 1.
  • So f is continuous at both points.
  • x = −1 पर: f(−1) = −2; बायाँ सीमा = −2; दायाँ सीमा = lim 2(−1 + h) = −2। सब बराबर ⇒ x = −1 पर संतत।
  • x = 1 पर: f(1) = 2(1) = 2; बायाँ सीमा = lim 2(1 − h) = 2; दायाँ सीमा = 2। सब बराबर ⇒ x = 1 पर संतत।
  • अतः f दोनों बिंदुओं पर संतत है।
  • x = −1 par: f(−1) = −2; LHL = −2; RHL = lim 2(−1 + h) = −2. Sab barabar ⇒ x = −1 par continuous.
  • x = 1 par: f(1) = 2(1) = 2; LHL = lim 2(1 − h) = 2; RHL = 2. Sab barabar ⇒ x = 1 par continuous.
  • Isliye f dono points par continuous hai.

💡 Write all three values (f(a), LHL, RHL) separately with their working; each carries a step mark.तीनों मान (f(a), बायाँ सीमा, दायाँ सीमा) अलग-अलग गणना सहित लिखिए; प्रत्येक के अंक हैं।Teeno values (f(a), LHL, RHL) alag-alag working ke saath likhiye; har ek ke step marks hain.

Second-order derivative: "show that" relationsद्वितीय कोटि अवकलज: "दिखाइए कि" संबंधSecond order derivative: "dikhaiye ki" relations

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  • Write y₁ = dy/dx and y₂ = d²y/dx² to save time.
  • Golden trick: after finding y₁, cross-multiply to remove fractions/roots, THEN differentiate again. This gives the required relation directly.
  • y = sin⁻¹x ⇒ √(1 − x²) y₁ = 1 ⇒ (1 − x²) y₂ − x y₁ = 0.
  • y = e^(tan⁻¹x) ⇒ (1 + x²) y₁ = y ⇒ (1 + x²) y₂ + (2x − 1) y₁ = 0.
  • y = Ae^(mx) + Be^(nx) ⇒ y₂ − (m + n) y₁ + mny = 0.
  • y = A cos(log x) + B sin(log x) ⇒ x² y₂ + x y₁ + y = 0.
  • y = (tan⁻¹x)² ⇒ (1 + x²)² y₂ + 2x(1 + x²) y₁ = 2.
  • समय बचाने के लिए y₁ = dy/dx तथा y₂ = d²y/dx² लिखिए।
  • सुनहरी युक्ति: y₁ निकालने के बाद भिन्न/मूल हटाने के लिए वज्र-गुणन कीजिए, फिर दोबारा अवकलन कीजिए। इससे सीधे अभीष्ट संबंध मिलता है।
  • y = sin⁻¹x ⇒ √(1 − x²) y₁ = 1 ⇒ (1 − x²) y₂ − x y₁ = 0।
  • y = e^(tan⁻¹x) ⇒ (1 + x²) y₁ = y ⇒ (1 + x²) y₂ + (2x − 1) y₁ = 0।
  • y = Ae^(mx) + Be^(nx) ⇒ y₂ − (m + n) y₁ + mny = 0।
  • y = A cos(log x) + B sin(log x) ⇒ x² y₂ + x y₁ + y = 0।
  • y = (tan⁻¹x)² ⇒ (1 + x²)² y₂ + 2x(1 + x²) y₁ = 2।
  • Time bachane ke liye y₁ = dy/dx aur y₂ = d²y/dx² likhiye.
  • Golden trick: y₁ nikalne ke baad fraction/root hatane ke liye cross-multiply kijiye, PHIR dobara differentiate kijiye. Isse seedha required relation milta hai.
  • y = sin⁻¹x ⇒ √(1 − x²) y₁ = 1 ⇒ (1 − x²) y₂ − x y₁ = 0.
  • y = e^(tan⁻¹x) ⇒ (1 + x²) y₁ = y ⇒ (1 + x²) y₂ + (2x − 1) y₁ = 0.
  • y = Ae^(mx) + Be^(nx) ⇒ y₂ − (m + n) y₁ + mny = 0.
  • y = A cos(log x) + B sin(log x) ⇒ x² y₂ + x y₁ + y = 0.
  • y = (tan⁻¹x)² ⇒ (1 + x²)² y₂ + 2x(1 + x²) y₁ = 2.
Repeated questionबार-बार पूछा गया प्रश्नBaar baar aaya sawaal · 5 marks5 अंक5 marks

If y = sin⁻¹x, show that (1 − x²) d²y/dx² − x dy/dx = 0.यदि y = sin⁻¹x, तो दिखाइए कि (1 − x²) d²y/dx² − x dy/dx = 0।Agar y = sin⁻¹x, to dikhaiye ki (1 − x²) d²y/dx² − x dy/dx = 0.

✅ Model answerआदर्श उत्तरModel answer
  • y₁ = 1/√(1 − x²) ⇒ √(1 − x²)·y₁ = 1.
  • Differentiate: √(1 − x²)·y₂ + y₁·(−2x)/(2√(1 − x²)) = 0.
  • Multiply by √(1 − x²): (1 − x²) y₂ − x y₁ = 0. Proved.
  • y₁ = 1/√(1 − x²) ⇒ √(1 − x²)·y₁ = 1।
  • अवकलन: √(1 − x²)·y₂ + y₁·(−2x)/(2√(1 − x²)) = 0।
  • √(1 − x²) से गुणा: (1 − x²) y₂ − x y₁ = 0। इति सिद्धम्।
  • y₁ = 1/√(1 − x²) ⇒ √(1 − x²)·y₁ = 1.
  • Differentiate: √(1 − x²)·y₂ + y₁·(−2x)/(2√(1 − x²)) = 0.
  • √(1 − x²) se multiply: (1 − x²) y₂ − x y₁ = 0. Proved.
Repeated questionबार-बार पूछा गया प्रश्नBaar baar aaya sawaal · 5 marks5 अंक5 marks

If y = e^(tan⁻¹x), prove that (1 + x²) d²y/dx² + (2x − 1) dy/dx = 0.यदि y = e^(tan⁻¹x), तो सिद्ध कीजिए कि (1 + x²) d²y/dx² + (2x − 1) dy/dx = 0।Agar y = e^(tan⁻¹x), to prove kijiye ki (1 + x²) d²y/dx² + (2x − 1) dy/dx = 0.

✅ Model answerआदर्श उत्तरModel answer
  • y₁ = e^(tan⁻¹x)·1/(1 + x²) = y/(1 + x²) ⇒ (1 + x²) y₁ = y.
  • Differentiate: (1 + x²) y₂ + 2x y₁ = y₁.
  • (1 + x²) y₂ + (2x − 1) y₁ = 0. Proved.
  • y₁ = e^(tan⁻¹x)·1/(1 + x²) = y/(1 + x²) ⇒ (1 + x²) y₁ = y।
  • अवकलन: (1 + x²) y₂ + 2x y₁ = y₁।
  • (1 + x²) y₂ + (2x − 1) y₁ = 0। इति सिद्धम्।
  • y₁ = e^(tan⁻¹x)·1/(1 + x²) = y/(1 + x²) ⇒ (1 + x²) y₁ = y.
  • Differentiate: (1 + x²) y₂ + 2x y₁ = y₁.
  • (1 + x²) y₂ + (2x − 1) y₁ = 0. Proved.
Repeated questionबार-बार पूछा गया प्रश्नBaar baar aaya sawaal · 5 marks5 अंक5 marks

Worked example: If y = 3e^(2x) + 2e^(3x), prove that d²y/dx² − 5 dy/dx + 6y = 0.हल किया उदाहरण: यदि y = 3e^(2x) + 2e^(3x), तो सिद्ध कीजिए कि d²y/dx² − 5 dy/dx + 6y = 0।Solved example: agar y = 3e^(2x) + 2e^(3x), to prove kijiye ki d²y/dx² − 5 dy/dx + 6y = 0.

✅ Model answerआदर्श उत्तरModel answer
  • y₁ = 6e^(2x) + 6e^(3x).
  • y₂ = 12e^(2x) + 18e^(3x).
  • y₂ − 5y₁ + 6y = (12 − 30 + 18)e^(2x) + (18 − 30 + 12)e^(3x) = 0 + 0 = 0. Proved.
  • (This is the case m = 2, n = 3 of y₂ − (m + n)y₁ + mny = 0.)
  • y₁ = 6e^(2x) + 6e^(3x)।
  • y₂ = 12e^(2x) + 18e^(3x)।
  • y₂ − 5y₁ + 6y = (12 − 30 + 18)e^(2x) + (18 − 30 + 12)e^(3x) = 0 + 0 = 0। इति सिद्धम्।
  • (यह y₂ − (m + n)y₁ + mny = 0 में m = 2, n = 3 की स्थिति है।)
  • y₁ = 6e^(2x) + 6e^(3x).
  • y₂ = 12e^(2x) + 18e^(3x).
  • y₂ − 5y₁ + 6y = (12 − 30 + 18)e^(2x) + (18 − 30 + 12)e^(3x) = 0 + 0 = 0. Proved.
  • (Yeh y₂ − (m + n)y₁ + mny = 0 mein m = 2, n = 3 wala case hai.)

💡 Never substitute y₁ = 1/√(1 − x²) into the final relation and simplify fractions; the cross-multiply-then-differentiate route is shorter and safer.अंतिम संबंध में y₁ = 1/√(1 − x²) रखकर भिन्नों का सरलीकरण मत कीजिए; वज्र-गुणन फिर अवकलन वाला रास्ता छोटा और सुरक्षित है।Final relation mein y₁ = 1/√(1 − x²) rakh kar fractions simplify mat kijiye; cross-multiply phir differentiate wala raasta chhota aur safe hai.

Logarithmic differentiation (x^x, x^(sin x) type)लघुगणकीय अवकलन (x^x, x^(sin x) प्रकार)Logarithmic differentiation (x^x, x^(sin x) type)

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  • Use it when the variable is in the power: [f(x)]^(g(x)).
  • For y = u + v (sum of two powers): NEVER take log of the whole sum. Put u and v separately, find du/dx and dv/dx, then dy/dx = du/dx + dv/dx.
  • For u = [f(x)]^(g(x)): log u = g(x)·log f(x) ⇒ (1/u) du/dx = g′(x) log f(x) + g(x)·f′(x)/f(x).
  • d/dx (x^x) = x^x (1 + log x).
  • d/dx (x^(sin x)) = x^(sin x) [cos x · log x + (sin x)/x].
  • d/dx ((cos x)^(tan x)) = (cos x)^(tan x) [sec²x · log cos x − tan²x].
  • Also used for long products/quotients, and for x^y = e^(x − y) ⇒ dy/dx = log x/(1 + log x)².
  • जब चर घात में हो तब इसका प्रयोग कीजिए: [f(x)]^(g(x))।
  • y = u + v (दो घातों का योग) के लिए पूरे योग का log कभी मत लीजिए। u तथा v अलग रखिए, du/dx और dv/dx निकालिए, फिर dy/dx = du/dx + dv/dx।
  • u = [f(x)]^(g(x)) के लिए: log u = g(x)·log f(x) ⇒ (1/u) du/dx = g′(x) log f(x) + g(x)·f′(x)/f(x)।
  • d/dx (x^x) = x^x (1 + log x)।
  • d/dx (x^(sin x)) = x^(sin x) [cos x · log x + (sin x)/x]।
  • d/dx ((cos x)^(tan x)) = (cos x)^(tan x) [sec²x · log cos x − tan²x]।
  • लंबे गुणनफल/भागफल में भी, तथा x^y = e^(x − y) ⇒ dy/dx = log x/(1 + log x)²।
  • Jab variable power mein ho tab ise use kijiye: [f(x)]^(g(x)).
  • y = u + v (do powers ka jod) ke liye poore sum ka log kabhi mat lijiye. u aur v alag rakhiye, du/dx aur dv/dx nikaliye, phir dy/dx = du/dx + dv/dx.
  • u = [f(x)]^(g(x)) ke liye: log u = g(x)·log f(x) ⇒ (1/u) du/dx = g′(x) log f(x) + g(x)·f′(x)/f(x).
  • d/dx (x^x) = x^x (1 + log x).
  • d/dx (x^(sin x)) = x^(sin x) [cos x · log x + (sin x)/x].
  • d/dx ((cos x)^(tan x)) = (cos x)^(tan x) [sec²x · log cos x − tan²x].
  • Lambe product/quotient mein bhi, aur x^y = e^(x − y) ⇒ dy/dx = log x/(1 + log x)².
✏️ Formulas to rememberयाद रखने वाले सूत्रYaad rakhne wale formulas
d/dx [f^g] = f^g [g′ log f + g f′/f]
d/dx (x^x) = x^x(1 + log x)
Repeated questionबार-बार पूछा गया प्रश्नBaar baar aaya sawaal · 5 marks5 अंक5 marks

Find dy/dx if y = x^(cos x) + (sin x)^x.यदि y = x^(cos x) + (sin x)^x, तो dy/dx ज्ञात कीजिए।Agar y = x^(cos x) + (sin x)^x, to dy/dx nikaliye.

✅ Model answerआदर्श उत्तरModel answer
  • Let u = x^(cos x), v = (sin x)^x; y = u + v.
  • log u = cos x · log x ⇒ (1/u) du/dx = −sin x · log x + cos x/x ⇒ du/dx = x^(cos x)[cos x/x − sin x log x].
  • log v = x · log sin x ⇒ (1/v) dv/dx = log sin x + x cot x ⇒ dv/dx = (sin x)^x [log sin x + x cot x].
  • dy/dx = x^(cos x)[cos x/x − sin x log x] + (sin x)^x [log sin x + x cot x].
  • माना u = x^(cos x), v = (sin x)^x; y = u + v।
  • log u = cos x · log x ⇒ (1/u) du/dx = −sin x · log x + cos x/x ⇒ du/dx = x^(cos x)[cos x/x − sin x log x]।
  • log v = x · log sin x ⇒ (1/v) dv/dx = log sin x + x cot x ⇒ dv/dx = (sin x)^x [log sin x + x cot x]।
  • dy/dx = x^(cos x)[cos x/x − sin x log x] + (sin x)^x [log sin x + x cot x]।
  • Maana u = x^(cos x), v = (sin x)^x; y = u + v.
  • log u = cos x · log x ⇒ (1/u) du/dx = −sin x · log x + cos x/x ⇒ du/dx = x^(cos x)[cos x/x − sin x log x].
  • log v = x · log sin x ⇒ (1/v) dv/dx = log sin x + x cot x ⇒ dv/dx = (sin x)^x [log sin x + x cot x].
  • dy/dx = x^(cos x)[cos x/x − sin x log x] + (sin x)^x [log sin x + x cot x].
Repeated questionबार-बार पूछा गया प्रश्नBaar baar aaya sawaal · 5 marks5 अंक5 marks

Worked example: Differentiate y = x^x − 2^(sin x).हल किया उदाहरण: y = x^x − 2^(sin x) का अवकलन कीजिए।Solved example: y = x^x − 2^(sin x) ka differentiation kijiye.

✅ Model answerआदर्श उत्तरModel answer
  • Let u = x^x: log u = x log x ⇒ (1/u) du/dx = log x + 1 ⇒ du/dx = x^x(1 + log x).
  • Let v = 2^(sin x): this is a^(f(x)) with a constant base, so dv/dx = 2^(sin x) · log 2 · cos x.
  • dy/dx = x^x(1 + log x) − 2^(sin x) cos x · log 2.
  • माना u = x^x: log u = x log x ⇒ (1/u) du/dx = log x + 1 ⇒ du/dx = x^x(1 + log x)।
  • माना v = 2^(sin x): आधार अचर है, अतः dv/dx = 2^(sin x) · log 2 · cos x।
  • dy/dx = x^x(1 + log x) − 2^(sin x) cos x · log 2।
  • Maana u = x^x: log u = x log x ⇒ (1/u) du/dx = log x + 1 ⇒ du/dx = x^x(1 + log x).
  • Maana v = 2^(sin x): base constant hai, isliye dv/dx = 2^(sin x) · log 2 · cos x.
  • dy/dx = x^x(1 + log x) − 2^(sin x) cos x · log 2.

Implicit functionsअस्पष्ट फलनImplicit functions

🔥 Asked in 5 of 7 years🔥 7 में से 5 वर्ष पूछा गया🔥 Aaya hai 5/7 saal🔴 Very importantबहुत महत्वपूर्णBahut important
Came / not askedआया / नहीं आयाAaya / nahi aaya: 2019202020222023202420252026
  • Differentiate both sides w.r.t. x; every time you differentiate a term in y, multiply by dy/dx.
  • Collect dy/dx terms on one side and solve.
  • Shortcut: for F(x, y) = 0, dy/dx = −(∂F/∂x)/(∂F/∂y) (use only to check).
  • cos y = x cos(a + y) ⇒ write x = cos y/cos(a + y), find dx/dy, then dy/dx = cos²(a + y)/sin a.
  • x√(1 + y) + y√(1 + x) = 0 ⇒ square and factorise to get y = −x/(1 + x), then dy/dx = −1/(1 + x)².
  • दोनों पक्षों का x के सापेक्ष अवकलन कीजिए; y वाले पद के अवकलन पर हर बार dy/dx से गुणा कीजिए।
  • dy/dx वाले पद एक ओर एकत्र करके हल कीजिए।
  • संक्षिप्त: F(x, y) = 0 के लिए dy/dx = −(∂F/∂x)/(∂F/∂y) (केवल जाँच के लिए)।
  • cos y = x cos(a + y) ⇒ x = cos y/cos(a + y) लिखिए, dx/dy निकालिए, फिर dy/dx = cos²(a + y)/sin a।
  • x√(1 + y) + y√(1 + x) = 0 ⇒ वर्ग करके गुणनखंड से y = −x/(1 + x), फिर dy/dx = −1/(1 + x)²।
  • Dono sides ka x ke saapeksh differentiation kijiye; y wale term ko differentiate karne par har baar dy/dx se multiply kijiye.
  • dy/dx wale terms ek side ikattha karke solve kijiye.
  • Shortcut: F(x, y) = 0 ke liye dy/dx = −(∂F/∂x)/(∂F/∂y) (sirf check ke liye).
  • cos y = x cos(a + y) ⇒ x = cos y/cos(a + y) likhiye, dx/dy nikaliye, phir dy/dx = cos²(a + y)/sin a.
  • x√(1 + y) + y√(1 + x) = 0 ⇒ square karke factorise se y = −x/(1 + x), phir dy/dx = −1/(1 + x)².
Repeated questionबार-बार पूछा गया प्रश्नBaar baar aaya sawaal · 5 marks5 अंक5 marks

If x√(1 + y) + y√(1 + x) = 0, −1 < x < 1, prove that dy/dx = −1/(1 + x)².यदि x√(1 + y) + y√(1 + x) = 0, −1 < x < 1, तो सिद्ध कीजिए कि dy/dx = −1/(1 + x)²।Agar x√(1 + y) + y√(1 + x) = 0, −1 < x < 1, to prove kijiye ki dy/dx = −1/(1 + x)².

✅ Model answerआदर्श उत्तरModel answer
  • x√(1 + y) = −y√(1 + x). Square: x²(1 + y) = y²(1 + x).
  • x² − y² + x²y − xy² = 0 ⇒ (x − y)(x + y) + xy(x − y) = 0 ⇒ (x − y)(x + y + xy) = 0.
  • x = y does not satisfy the given equation (it gives 2x√(1 + x) = 0 only at x = 0), so x + y + xy = 0.
  • y(1 + x) = −x ⇒ y = −x/(1 + x).
  • dy/dx = −[(1 + x)·1 − x·1]/(1 + x)² = −1/(1 + x)². Proved.
  • x√(1 + y) = −y√(1 + x)। वर्ग: x²(1 + y) = y²(1 + x)।
  • x² − y² + x²y − xy² = 0 ⇒ (x − y)(x + y) + xy(x − y) = 0 ⇒ (x − y)(x + y + xy) = 0।
  • x = y दिए समीकरण को संतुष्ट नहीं करता (इससे 2x√(1 + x) = 0 केवल x = 0 पर), अतः x + y + xy = 0।
  • y(1 + x) = −x ⇒ y = −x/(1 + x)।
  • dy/dx = −[(1 + x)·1 − x·1]/(1 + x)² = −1/(1 + x)²। इति सिद्धम्।
  • x√(1 + y) = −y√(1 + x). Square: x²(1 + y) = y²(1 + x).
  • x² − y² + x²y − xy² = 0 ⇒ (x − y)(x + y) + xy(x − y) = 0 ⇒ (x − y)(x + y + xy) = 0.
  • x = y diye equation ko satisfy nahi karta (isse 2x√(1 + x) = 0 sirf x = 0 par), isliye x + y + xy = 0.
  • y(1 + x) = −x ⇒ y = −x/(1 + x).
  • dy/dx = −[(1 + x)·1 − x·1]/(1 + x)² = −1/(1 + x)². Proved.
Repeated questionबार-बार पूछा गया प्रश्नBaar baar aaya sawaal · 2 marks2 अंक2 marks

Worked example: If x² + xy + y² = 7, find dy/dx at (1, 2).हल किया उदाहरण: यदि x² + xy + y² = 7, तो (1, 2) पर dy/dx ज्ञात कीजिए।Solved example: agar x² + xy + y² = 7, to (1, 2) par dy/dx nikaliye.

✅ Model answerआदर्श उत्तरModel answer
  • Differentiate: 2x + (y + x dy/dx) + 2y dy/dx = 0.
  • dy/dx (x + 2y) = −(2x + y) ⇒ dy/dx = −(2x + y)/(x + 2y).
  • At (1, 2): dy/dx = −(2 + 2)/(1 + 4) = −4/5.
  • अवकलन: 2x + (y + x dy/dx) + 2y dy/dx = 0।
  • dy/dx (x + 2y) = −(2x + y) ⇒ dy/dx = −(2x + y)/(x + 2y)।
  • (1, 2) पर: dy/dx = −(2 + 2)/(1 + 4) = −4/5।
  • Differentiate: 2x + (y + x dy/dx) + 2y dy/dx = 0.
  • dy/dx (x + 2y) = −(2x + y) ⇒ dy/dx = −(2x + y)/(x + 2y).
  • (1, 2) par: dy/dx = −(2 + 2)/(1 + 4) = −4/5.

Differentiability at a pointकिसी बिंदु पर अवकलनीयताPoint par differentiability

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  • f is differentiable at a if LHD = RHD, where LHD = lim(h→0) [f(a − h) − f(a)]/(−h) and RHD = lim(h→0) [f(a + h) − f(a)]/h.
  • Every differentiable function is continuous, but a continuous function need not be differentiable.
  • |x − a| is continuous at x = a but not differentiable there (LHD = −1, RHD = +1): a sharp corner.
  • Standard 5–8 mark question: show f(x) = |x − 2| is continuous but not differentiable at x = 2.
  • f, a पर अवकलनीय है यदि बायाँ अवकलज = दायाँ अवकलज, जहाँ बायाँ = lim(h→0) [f(a − h) − f(a)]/(−h) तथा दायाँ = lim(h→0) [f(a + h) − f(a)]/h।
  • प्रत्येक अवकलनीय फलन संतत होता है, परन्तु संतत फलन का अवकलनीय होना आवश्यक नहीं।
  • |x − a|, x = a पर संतत है परन्तु अवकलनीय नहीं (बायाँ = −1, दायाँ = +1): नुकीला कोना।
  • मानक 5–8 अंक का प्रश्न: दिखाइए f(x) = |x − 2|, x = 2 पर संतत है परन्तु अवकलनीय नहीं।
  • f, a par differentiable hai agar LHD = RHD, jahan LHD = lim(h→0) [f(a − h) − f(a)]/(−h) aur RHD = lim(h→0) [f(a + h) − f(a)]/h.
  • Har differentiable function continuous hota hai, lekin continuous function ka differentiable hona zaroori nahi.
  • |x − a|, x = a par continuous hai lekin differentiable nahi (LHD = −1, RHD = +1): nukila kona.
  • Standard 5–8 marks ka sawaal: dikhaiye f(x) = |x − 2|, x = 2 par continuous hai lekin differentiable nahi.
Repeated questionबार-बार पूछा गया प्रश्नBaar baar aaya sawaal · 5 marks5 अंक5 marks

Show that f(x) = |x − 2| is continuous but not differentiable at x = 2.दिखाइए कि f(x) = |x − 2|, x = 2 पर संतत है परन्तु अवकलनीय नहीं है।Dikhaiye ki f(x) = |x − 2|, x = 2 par continuous hai lekin differentiable nahi hai.

✅ Model answerआदर्श उत्तरModel answer
  • f(2) = 0. LHL = lim |2 − h − 2| = lim h = 0; RHL = lim |2 + h − 2| = lim h = 0. So f is continuous at x = 2.
  • LHD = lim(h→0) [f(2 − h) − f(2)]/(−h) = lim h/(−h) = −1.
  • RHD = lim(h→0) [f(2 + h) − f(2)]/h = lim h/h = 1.
  • LHD ≠ RHD, so f is not differentiable at x = 2.
  • f(2) = 0। बायाँ सीमा = lim |2 − h − 2| = lim h = 0; दायाँ सीमा = lim |2 + h − 2| = lim h = 0। अतः f, x = 2 पर संतत है।
  • बायाँ अवकलज = lim(h→0) [f(2 − h) − f(2)]/(−h) = lim h/(−h) = −1।
  • दायाँ अवकलज = lim(h→0) [f(2 + h) − f(2)]/h = lim h/h = 1।
  • बायाँ ≠ दायाँ, अतः f, x = 2 पर अवकलनीय नहीं है।
  • f(2) = 0. LHL = lim |2 − h − 2| = lim h = 0; RHL = lim |2 + h − 2| = lim h = 0. Isliye f, x = 2 par continuous hai.
  • LHD = lim(h→0) [f(2 − h) − f(2)]/(−h) = lim h/(−h) = −1.
  • RHD = lim(h→0) [f(2 + h) − f(2)]/h = lim h/h = 1.
  • LHD ≠ RHD, isliye f, x = 2 par differentiable nahi hai.

Derivatives of inverse trig functions (with substitution)प्रतिलोम त्रिकोणमितीय फलनों के अवकलज (प्रतिस्थापन सहित)Inverse trig functions ke derivatives (substitution ke saath)

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  • 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)) (x > 1).
  • Simplify first, then differentiate: tan⁻¹[sin x/(1 + cos x)] = tan⁻¹[tan(x/2)] = x/2, so the derivative is 1/2.
  • 2x/(1 − x²), 2x/(1 + x²), (1 − x²)/(1 + x²) → put x = tan θ; each becomes 2tan⁻¹x on the right interval.
  • Derivative of u w.r.t. v: du/dv = (du/dx)/(dv/dx). E.g. tan⁻¹[2x/(1 − x²)] w.r.t. cos⁻¹[(1 − x²)/(1 + x²)] = 1 for 0 < x < 1.
  • 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)) (x > 1)।
  • पहले सरल कीजिए, फिर अवकलन: tan⁻¹[sin x/(1 + cos x)] = tan⁻¹[tan(x/2)] = x/2, अतः अवकलज 1/2।
  • 2x/(1 − x²), 2x/(1 + x²), (1 − x²)/(1 + x²) → x = tan θ रखिए; उचित अंतराल पर प्रत्येक 2tan⁻¹x बनता है।
  • u का v के सापेक्ष अवकलज: du/dv = (du/dx)/(dv/dx)। जैसे 0 < x < 1 पर tan⁻¹[2x/(1 − x²)] का cos⁻¹[(1 − x²)/(1 + x²)] के सापेक्ष अवकलज = 1।
  • 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)) (x > 1).
  • Pehle simplify, phir differentiate: tan⁻¹[sin x/(1 + cos x)] = tan⁻¹[tan(x/2)] = x/2, isliye derivative 1/2.
  • 2x/(1 − x²), 2x/(1 + x²), (1 − x²)/(1 + x²) → x = tan θ rakhiye; sahi interval par har ek 2tan⁻¹x banta hai.
  • u ka v ke saapeksh derivative: du/dv = (du/dx)/(dv/dx). Jaise 0 < x < 1 par tan⁻¹[2x/(1 − x²)] ka cos⁻¹[(1 − x²)/(1 + x²)] ke saapeksh derivative = 1.
Repeated questionबार-बार पूछा गया प्रश्नBaar baar aaya sawaal · 5 marks5 अंक5 marks

Differentiate tan⁻¹[sin x/(1 + cos x)] with respect to x.tan⁻¹[sin x/(1 + cos x)] का x के सापेक्ष अवकलन कीजिए।tan⁻¹[sin x/(1 + cos x)] ka x ke saapeksh differentiation kijiye.

✅ Model answerआदर्श उत्तरModel answer
  • sin x = 2 sin(x/2) cos(x/2) and 1 + cos x = 2cos²(x/2).
  • sin x/(1 + cos x) = tan(x/2).
  • y = tan⁻¹[tan(x/2)] = x/2 (for −π < x < π).
  • dy/dx = 1/2.
  • sin x = 2 sin(x/2) cos(x/2) तथा 1 + cos x = 2cos²(x/2)।
  • sin x/(1 + cos x) = tan(x/2)।
  • y = tan⁻¹[tan(x/2)] = x/2 (−π < x < π के लिए)।
  • dy/dx = 1/2।
  • sin x = 2 sin(x/2) cos(x/2) aur 1 + cos x = 2cos²(x/2).
  • sin x/(1 + cos x) = tan(x/2).
  • y = tan⁻¹[tan(x/2)] = x/2 (−π < x < π ke liye).
  • dy/dx = 1/2.

Continuity of standard functionsमानक फलनों का सांतत्यStandard functions ki continuity

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  • Polynomials, sin x, cos x, eˣ and |x| are continuous everywhere.
  • tan x, sec x, log x, rational functions are continuous at every point of their domain (where they are defined).
  • Sum, difference, product and quotient (denominator ≠ 0) of continuous functions are continuous; so is a composition.
  • |x| is continuous: |x| = −x for x < 0 and x for x ≥ 0; at 0, LHL = RHL = f(0) = 0.
  • The greatest integer function [x] is discontinuous at every integer.
  • बहुपद, sin x, cos x, eˣ तथा |x| सर्वत्र संतत हैं।
  • tan x, sec x, log x, परिमेय फलन अपने प्रांत के प्रत्येक बिंदु पर संतत हैं।
  • संतत फलनों का योग, अंतर, गुणनफल तथा भागफल (हर ≠ 0) संतत होता है; संयोजन भी।
  • |x| संतत है: x < 0 के लिए |x| = −x तथा x ≥ 0 के लिए x; 0 पर बायाँ सीमा = दायाँ सीमा = f(0) = 0।
  • महत्तम पूर्णांक फलन [x] प्रत्येक पूर्णांक पर असंतत है।
  • Polynomials, sin x, cos x, eˣ aur |x| har jagah continuous hain.
  • tan x, sec x, log x, rational functions apne domain ke har point par continuous hain.
  • Continuous functions ka sum, difference, product aur quotient (denominator ≠ 0) continuous hota hai; composition bhi.
  • |x| continuous hai: x < 0 ke liye |x| = −x aur x ≥ 0 ke liye x; 0 par LHL = RHL = f(0) = 0.
  • Greatest integer function [x] har integer par discontinuous hai.
Repeated questionबार-बार पूछा गया प्रश्नBaar baar aaya sawaal · 2 marks2 अंक2 marks

Prove that f(x) = tan x is a continuous function.सिद्ध कीजिए कि f(x) = tan x एक संतत फलन है।Prove kijiye ki f(x) = tan x ek continuous function hai.

✅ Model answerआदर्श उत्तरModel answer
  • tan x = sin x/cos x, defined for all x except odd multiples of π/2 (where cos x = 0).
  • sin x and cos x are continuous everywhere.
  • A quotient of continuous functions is continuous wherever the denominator ≠ 0.
  • So tan x is continuous at every point of its domain, i.e. it is a continuous function.
  • tan x = sin x/cos x, π/2 के विषम गुणजों (जहाँ cos x = 0) को छोड़कर सभी x के लिए परिभाषित।
  • sin x तथा cos x सर्वत्र संतत हैं।
  • संतत फलनों का भागफल वहाँ संतत है जहाँ हर ≠ 0।
  • अतः tan x अपने प्रांत के प्रत्येक बिंदु पर संतत है, अर्थात संतत फलन है।
  • tan x = sin x/cos x, π/2 ke odd multiples (jahan cos x = 0) ko chhod kar saare x ke liye defined.
  • sin x aur cos x har jagah continuous hain.
  • Continuous functions ka quotient wahan continuous hai jahan denominator ≠ 0.
  • Isliye tan x apne domain ke har point par continuous hai, yaani continuous function hai.

Parametric differentiationप्राचलिक अवकलनParametric differentiation

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  • If x = f(t), y = g(t): dy/dx = (dy/dt)/(dx/dt).
  • Second derivative: d²y/dx² = d/dt(dy/dx) ÷ (dx/dt). Do NOT write (d²y/dt²)/(d²x/dt²).
  • x = at², y = 2at ⇒ dy/dx = 1/t, d²y/dx² = −1/(2at³).
  • x = a(θ + sin θ), y = a(1 − cos θ) ⇒ dy/dx = sin θ/(1 + cos θ) = tan(θ/2).
  • x = a sin³t, y = b cos³t ⇒ dy/dx = −(b/a) cot t; at t = π/2, dy/dx = 0.
  • यदि x = f(t), y = g(t): dy/dx = (dy/dt)/(dx/dt)।
  • द्वितीय अवकलज: d²y/dx² = d/dt(dy/dx) ÷ (dx/dt)। (d²y/dt²)/(d²x/dt²) मत लिखिए।
  • x = at², y = 2at ⇒ dy/dx = 1/t, d²y/dx² = −1/(2at³)।
  • x = a(θ + sin θ), y = a(1 − cos θ) ⇒ dy/dx = sin θ/(1 + cos θ) = tan(θ/2)।
  • x = a sin³t, y = b cos³t ⇒ dy/dx = −(b/a) cot t; t = π/2 पर dy/dx = 0।
  • Agar x = f(t), y = g(t): dy/dx = (dy/dt)/(dx/dt).
  • Second derivative: d²y/dx² = d/dt(dy/dx) ÷ (dx/dt). (d²y/dt²)/(d²x/dt²) mat likhiye.
  • x = at², y = 2at ⇒ dy/dx = 1/t, d²y/dx² = −1/(2at³).
  • x = a(θ + sin θ), y = a(1 − cos θ) ⇒ dy/dx = sin θ/(1 + cos θ) = tan(θ/2).
  • x = a sin³t, y = b cos³t ⇒ dy/dx = −(b/a) cot t; t = π/2 par dy/dx = 0.
✏️ Formulas to rememberयाद रखने वाले सूत्रYaad rakhne wale formulas
dy/dx = (dy/dt)/(dx/dt)
d²y/dx² = [d/dt (dy/dx)] / (dx/dt)
Repeated questionबार-बार पूछा गया प्रश्नBaar baar aaya sawaal · 4 marks4 अंक4 marks

If x = a(cos t + t sin t) and y = a(sin t − t cos t), find d²y/dx².यदि x = a(cos t + t sin t) तथा y = a(sin t − t cos t), तो d²y/dx² ज्ञात कीजिए।Agar x = a(cos t + t sin t) aur y = a(sin t − t cos t), to d²y/dx² nikaliye.

✅ Model answerआदर्श उत्तरModel answer
  • 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/dt(tan t) ÷ dx/dt = sec²t/(at cos t) = sec³t/(at).
  • 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/dt(tan t) ÷ dx/dt = sec²t/(at cos t) = sec³t/(at)।
  • 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/dt(tan t) ÷ dx/dt = sec²t/(at cos t) = sec³t/(at).

Chain ruleशृंखला नियमChain rule

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  • d/dx f(g(x)) = f′(g(x))·g′(x): differentiate the outer function, keep the inside, multiply by the derivative of the inside.
  • d/dx sin(x² + 5) = 2x cos(x² + 5).
  • d/dx log(log x) = 1/(x log x), x > 1.
  • Derivative of sin²x w.r.t. e^(cos x) = (2 sin x cos x)/(−sin x e^(cos x)) = −2cos x/e^(cos x).
  • y = √[(1 − x)/(1 + x)] ⇒ (1 − x²) dy/dx + y = 0.
  • d/dx f(g(x)) = f′(g(x))·g′(x): बाहरी फलन का अवकलन, अंदर वाला वैसा ही, फिर अंदर वाले के अवकलज से गुणा।
  • d/dx sin(x² + 5) = 2x cos(x² + 5)।
  • d/dx log(log x) = 1/(x log x), x > 1।
  • sin²x का e^(cos x) के सापेक्ष अवकलज = (2 sin x cos x)/(−sin x e^(cos x)) = −2cos x/e^(cos x)।
  • y = √[(1 − x)/(1 + x)] ⇒ (1 − x²) dy/dx + y = 0।
  • d/dx f(g(x)) = f′(g(x))·g′(x): bahar wale function ka derivative, andar wala waisa hi, phir andar wale ke derivative se multiply.
  • d/dx sin(x² + 5) = 2x cos(x² + 5).
  • d/dx log(log x) = 1/(x log x), x > 1.
  • sin²x ka e^(cos x) ke saapeksh derivative = (2 sin x cos x)/(−sin x e^(cos x)) = −2cos x/e^(cos x).
  • y = √[(1 − x)/(1 + x)] ⇒ (1 − x²) dy/dx + y = 0.

Derivatives of exponential and log functionsचरघातांकी तथा लघुगणकीय फलनों के अवकलजExponential aur log functions ke derivatives

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  • d/dx eˣ = eˣ; d/dx aˣ = aˣ log a; d/dx e^(f(x)) = e^(f(x))·f′(x).
  • d/dx log x = 1/x (x > 0); d/dx log f(x) = f′(x)/f(x).
  • y = (x − 1) log(x − 1) − (x + 1) log(x + 1) ⇒ dy/dx = log[(x − 1)/(x + 1)] (the +1 and −1 cancel).
  • d/dx eˣ = eˣ; d/dx aˣ = aˣ log a; d/dx e^(f(x)) = e^(f(x))·f′(x)।
  • d/dx log x = 1/x (x > 0); d/dx log f(x) = f′(x)/f(x)।
  • y = (x − 1) log(x − 1) − (x + 1) log(x + 1) ⇒ dy/dx = log[(x − 1)/(x + 1)] (+1 तथा −1 कट जाते हैं)।
  • d/dx eˣ = eˣ; d/dx aˣ = aˣ log a; d/dx e^(f(x)) = e^(f(x))·f′(x).
  • d/dx log x = 1/x (x > 0); d/dx log f(x) = f′(x)/f(x).
  • y = (x − 1) log(x − 1) − (x + 1) log(x + 1) ⇒ dy/dx = log[(x − 1)/(x + 1)] (+1 aur −1 cancel ho jaate hain).
🔴 TOPPERटॉपरTOPPER

Full derivations, tricky questions, common mistakes and quick revision.पूरी व्युत्पत्तियाँ, कठिन प्रश्न, आम गलतियाँ और त्वरित पुनरावृत्ति।Poori derivations, tricky sawaal, aam galtiyan aur quick revision.

📐 Prove: every differentiable function is continuousसिद्ध कीजिए: प्रत्येक अवकलनीय फलन संतत होता हैProve kijiye: har differentiable function continuous hota hai

  1. Let f be differentiable at x = c, so f′(c) = lim(x→c) [f(x) − f(c)]/(x − c) exists.
  2. For x ≠ c: f(x) − f(c) = {[f(x) − f(c)]/(x − c)} · (x − c).
  3. Take the limit: lim(x→c) [f(x) − f(c)] = f′(c) · lim(x→c)(x − c) = f′(c) · 0 = 0.
  4. So lim(x→c) f(x) = f(c), i.e. f is continuous at c.
  5. The converse is false: |x| is continuous at 0 but not differentiable there.
  1. माना f, x = c पर अवकलनीय है, अतः f′(c) = lim(x→c) [f(x) − f(c)]/(x − c) विद्यमान है।
  2. x ≠ c के लिए: f(x) − f(c) = {[f(x) − f(c)]/(x − c)} · (x − c)।
  3. सीमा लेने पर: lim(x→c) [f(x) − f(c)] = f′(c) · lim(x→c)(x − c) = f′(c) · 0 = 0।
  4. अतः lim(x→c) f(x) = f(c), अर्थात f, c पर संतत है।
  5. विलोम सत्य नहीं: |x|, 0 पर संतत है परन्तु अवकलनीय नहीं।
  1. Maana f, x = c par differentiable hai, to f′(c) = lim(x→c) [f(x) − f(c)]/(x − c) exist karta hai.
  2. x ≠ c ke liye: f(x) − f(c) = {[f(x) − f(c)]/(x − c)} · (x − c).
  3. Limit lene par: lim(x→c) [f(x) − f(c)] = f′(c) · lim(x→c)(x − c) = f′(c) · 0 = 0.
  4. To lim(x→c) f(x) = f(c), yaani f, c par continuous hai.
  5. Ulta sach nahi: |x|, 0 par continuous hai lekin differentiable nahi.

📐 If cos y = x cos(a + y), cos a ≠ ±1, prove dy/dx = cos²(a + y)/sin aयदि cos y = x cos(a + y), cos a ≠ ±1, तो सिद्ध कीजिए dy/dx = cos²(a + y)/sin aAgar cos y = x cos(a + y), cos a ≠ ±1, to prove kijiye dy/dx = cos²(a + y)/sin a

  1. x = cos y/cos(a + y).
  2. dx/dy = [−sin y · cos(a + y) + cos y · sin(a + y)]/cos²(a + y).
  3. Numerator = sin(a + y)cos y − cos(a + y)sin y = sin[(a + y) − y] = sin a.
  4. dx/dy = sin a/cos²(a + y) ⇒ dy/dx = cos²(a + y)/sin a. Proved.
  1. x = cos y/cos(a + y)।
  2. dx/dy = [−sin y · cos(a + y) + cos y · sin(a + y)]/cos²(a + y)।
  3. अंश = sin(a + y)cos y − cos(a + y)sin y = sin[(a + y) − y] = sin a।
  4. dx/dy = sin a/cos²(a + y) ⇒ dy/dx = cos²(a + y)/sin a। इति सिद्धम्।
  1. x = cos y/cos(a + y).
  2. dx/dy = [−sin y · cos(a + y) + cos y · sin(a + y)]/cos²(a + y).
  3. Numerator = sin(a + y)cos y − cos(a + y)sin y = sin[(a + y) − y] = sin a.
  4. dx/dy = sin a/cos²(a + y) ⇒ dy/dx = cos²(a + y)/sin a. Proved.
🧠 Tricky questionकठिन प्रश्नTricky sawaal · 5 marks5 अंक5 marks

For what value of λ is f(x) = λ(x² − 2x) for x ≤ 0 and 4x + 1 for x > 0 continuous at x = 0?λ के किस मान के लिए f(x) = λ(x² − 2x), x ≤ 0 तथा 4x + 1, x > 0, x = 0 पर संतत है?λ ki kis value ke liye f(x) = λ(x² − 2x), x ≤ 0 aur 4x + 1, x > 0, x = 0 par continuous hai?

✅ Answerउत्तरJawab
  • f(0) = λ(0) = 0; LHL = lim λ[(−h)² + 2h] = 0.
  • RHL = lim [4h + 1] = 1.
  • LHL = 0 ≠ 1 = RHL whatever λ is.
  • So there is NO value of λ for which f is continuous at x = 0.
  • f(0) = λ(0) = 0; बायाँ सीमा = lim λ[(−h)² + 2h] = 0।
  • दायाँ सीमा = lim [4h + 1] = 1।
  • λ कुछ भी हो, बायाँ सीमा = 0 ≠ 1 = दायाँ सीमा।
  • अतः λ का कोई मान नहीं जिसके लिए f, x = 0 पर संतत हो।
  • f(0) = λ(0) = 0; LHL = lim λ[(−h)² + 2h] = 0.
  • RHL = lim [4h + 1] = 1.
  • λ kuch bhi ho, LHL = 0 ≠ 1 = RHL.
  • Isliye λ ki koi value nahi jiske liye f, x = 0 par continuous ho.
🧠 Tricky questionकठिन प्रश्नTricky sawaal · 5 marks5 अंक5 marks

If x^y = e^(x − y), prove that dy/dx = log x/(1 + log x)².यदि x^y = e^(x − y), तो सिद्ध कीजिए कि dy/dx = log x/(1 + log x)²।Agar x^y = e^(x − y), to prove kijiye ki dy/dx = log x/(1 + log x)².

✅ Answerउत्तरJawab
  • Take log: y log x = x − y ⇒ y(1 + log x) = x ⇒ y = x/(1 + log x).
  • dy/dx = [(1 + log x)·1 − x·(1/x)]/(1 + log x)².
  • = [1 + log x − 1]/(1 + log x)² = log x/(1 + log x)². Proved.
  • log लेने पर: y log x = x − y ⇒ y(1 + log x) = x ⇒ y = x/(1 + log x)।
  • dy/dx = [(1 + log x)·1 − x·(1/x)]/(1 + log x)²।
  • = [1 + log x − 1]/(1 + log x)² = log x/(1 + log x)²। इति सिद्धम्।
  • log lene par: y log x = x − y ⇒ y(1 + log x) = x ⇒ y = x/(1 + log x).
  • dy/dx = [(1 + log x)·1 − x·(1/x)]/(1 + log x)².
  • = [1 + log x − 1]/(1 + log x)² = log x/(1 + log x)². Proved.
🧠 Tricky questionकठिन प्रश्नTricky sawaal · 5 marks5 अंक5 marks

If y = (tan⁻¹x)², show that (1 + x²)² y₂ + 2x(1 + x²) y₁ = 2.यदि y = (tan⁻¹x)², तो दिखाइए कि (1 + x²)² y₂ + 2x(1 + x²) y₁ = 2।Agar y = (tan⁻¹x)², to dikhaiye ki (1 + x²)² y₂ + 2x(1 + x²) y₁ = 2.

✅ Answerउत्तरJawab
  • y₁ = 2 tan⁻¹x/(1 + x²) ⇒ (1 + x²) y₁ = 2 tan⁻¹x.
  • Differentiate: (1 + x²) y₂ + 2x y₁ = 2/(1 + x²).
  • Multiply by (1 + x²): (1 + x²)² y₂ + 2x(1 + x²) y₁ = 2. Proved.
  • y₁ = 2 tan⁻¹x/(1 + x²) ⇒ (1 + x²) y₁ = 2 tan⁻¹x।
  • अवकलन: (1 + x²) y₂ + 2x y₁ = 2/(1 + x²)।
  • (1 + x²) से गुणा: (1 + x²)² y₂ + 2x(1 + x²) y₁ = 2। इति सिद्धम्।
  • y₁ = 2 tan⁻¹x/(1 + x²) ⇒ (1 + x²) y₁ = 2 tan⁻¹x.
  • Differentiate: (1 + x²) y₂ + 2x y₁ = 2/(1 + x²).
  • (1 + x²) se multiply: (1 + x²)² y₂ + 2x(1 + x²) y₁ = 2. Proved.

⚠️ Common mistakesआम गलतियाँAam galtiyan

  • Taking log of a sum (log(u + v)) → split y = u + v and take log of u and v separately.
  • d/dx (x^x) = x·x^(x−1) → wrong; it is x^x(1 + log x).
  • d²y/dx² = (d²y/dt²)/(d²x/dt²) → wrong; use d/dt(dy/dx) ÷ dx/dt.
  • Checking only LHL = RHL → continuity also needs f(a).
  • Forgetting dy/dx when differentiating y² or sin y in implicit questions.
  • योग का log (log(u + v)) लेना → y = u + v तोड़कर u तथा v का अलग-अलग log लीजिए।
  • d/dx (x^x) = x·x^(x−1) → गलत; सही है x^x(1 + log x)।
  • d²y/dx² = (d²y/dt²)/(d²x/dt²) → गलत; d/dt(dy/dx) ÷ dx/dt लीजिए।
  • केवल बायाँ सीमा = दायाँ सीमा जाँचना → सांतत्य के लिए f(a) भी चाहिए।
  • अस्पष्ट प्रश्नों में y² या sin y के अवकलन में dy/dx भूल जाना।
  • Sum ka log (log(u + v)) lena → y = u + v todkar u aur v ka alag-alag log lijiye.
  • d/dx (x^x) = x·x^(x−1) → galat; sahi hai x^x(1 + log x).
  • d²y/dx² = (d²y/dt²)/(d²x/dt²) → galat; d/dt(dy/dx) ÷ dx/dt lijiye.
  • Sirf LHL = RHL check karna → continuity ke liye f(a) bhi chahiye.
  • Implicit sawaalon mein y² ya sin y ko differentiate karte waqt dy/dx bhool jaana.

💎 Rarely asked but importantकम पूछा गया पर ज़रूरीKam aaya par zaroori

  • x² sin(1/x) (with f(0) = 0) is continuous at 0 by the squeeze |x² sin(1/x)| ≤ x².
  • For ax + 1 (x ≤ 3) and bx + 3 (x > 3) to be continuous at 3: 3a + 1 = 3b + 3, i.e. a − b = 2/3 (many pairs work).
  • y = A cos(log x) + B sin(log x) ⇒ x²y₂ + xy₁ + y = 0.
  • Derivative of u w.r.t. v = (du/dx) ÷ (dv/dx).
  • x² sin(1/x) (f(0) = 0 के साथ) सैंडविच |x² sin(1/x)| ≤ x² से 0 पर संतत है।
  • x = 3 पर ax + 1 (x ≤ 3) तथा bx + 3 (x > 3) के सांतत्य के लिए: 3a + 1 = 3b + 3, अर्थात a − b = 2/3 (अनेक युग्म)।
  • y = A cos(log x) + B sin(log x) ⇒ x²y₂ + xy₁ + y = 0।
  • u का v के सापेक्ष अवकलज = (du/dx) ÷ (dv/dx)।
  • x² sin(1/x) (f(0) = 0 ke saath) squeeze |x² sin(1/x)| ≤ x² se 0 par continuous hai.
  • x = 3 par ax + 1 (x ≤ 3) aur bx + 3 (x > 3) ki continuity ke liye: 3a + 1 = 3b + 3, yaani a − b = 2/3 (kai pairs).
  • y = A cos(log x) + B sin(log x) ⇒ x²y₂ + xy₁ + y = 0.
  • u ka v ke saapeksh derivative = (du/dx) ÷ (dv/dx).

⚡ Quick revisionत्वरित पुनरावृत्तिQuick revision

  • Continuous at a: LHL = RHL = f(a).
  • Differentiable at a: LHD = RHD; differentiable ⇒ continuous.
  • d/dx xⁿ = nxⁿ⁻¹; sin → cos; cos → −sin; tan → sec²; sec → sec tan.
  • d/dx eˣ = eˣ; aˣ → aˣ log a; log x → 1/x.
  • sin⁻¹x → 1/√(1 − x²); tan⁻¹x → 1/(1 + x²); sec⁻¹x → 1/(x√(x² − 1)).
  • Chain: d/dx f(g(x)) = f′(g(x))·g′(x).
  • Parametric: dy/dx = (dy/dt)/(dx/dt).
  • x^x → x^x(1 + log x).
  • y = sin⁻¹x ⇒ (1 − x²)y₂ − xy₁ = 0.
  • y = Ae^(mx) + Be^(nx) ⇒ y₂ − (m + n)y₁ + mny = 0.
  • y = e^(tan⁻¹x) ⇒ (1 + x²)y₂ + (2x − 1)y₁ = 0.
  • |x − a|: continuous, not differentiable at a.
  • a पर संतत: बायाँ सीमा = दायाँ सीमा = f(a)।
  • a पर अवकलनीय: बायाँ अवकलज = दायाँ अवकलज; अवकलनीय ⇒ संतत।
  • d/dx xⁿ = nxⁿ⁻¹; sin → cos; cos → −sin; tan → sec²; sec → sec tan।
  • d/dx eˣ = eˣ; aˣ → aˣ log a; log x → 1/x।
  • sin⁻¹x → 1/√(1 − x²); tan⁻¹x → 1/(1 + x²); sec⁻¹x → 1/(x√(x² − 1))।
  • शृंखला: d/dx f(g(x)) = f′(g(x))·g′(x)।
  • प्राचलिक: dy/dx = (dy/dt)/(dx/dt)।
  • x^x → x^x(1 + log x)।
  • y = sin⁻¹x ⇒ (1 − x²)y₂ − xy₁ = 0।
  • y = Ae^(mx) + Be^(nx) ⇒ y₂ − (m + n)y₁ + mny = 0।
  • y = e^(tan⁻¹x) ⇒ (1 + x²)y₂ + (2x − 1)y₁ = 0।
  • |x − a|: संतत, a पर अवकलनीय नहीं।
  • a par continuous: LHL = RHL = f(a).
  • a par differentiable: LHD = RHD; differentiable ⇒ continuous.
  • d/dx xⁿ = nxⁿ⁻¹; sin → cos; cos → −sin; tan → sec²; sec → sec tan.
  • d/dx eˣ = eˣ; aˣ → aˣ log a; log x → 1/x.
  • sin⁻¹x → 1/√(1 − x²); tan⁻¹x → 1/(1 + x²); sec⁻¹x → 1/(x√(x² − 1)).
  • Chain: d/dx f(g(x)) = f′(g(x))·g′(x).
  • Parametric: dy/dx = (dy/dt)/(dx/dt).
  • x^x → x^x(1 + log x).
  • y = sin⁻¹x ⇒ (1 − x²)y₂ − xy₁ = 0.
  • y = Ae^(mx) + Be^(nx) ⇒ y₂ − (m + n)y₁ + mny = 0.
  • y = e^(tan⁻¹x) ⇒ (1 + x²)y₂ + (2x − 1)y₁ = 0.
  • |x − a|: continuous, a par differentiable nahi.

🚫 Removed, skip itहटाया गया, इसे छोड़ देंHata diya gaya, ise chhod dijiye

  • Rolle's theorem and the Mean Value Theorem (e.g. 'verify MVT for x³ − 5x² − 3x on [1, 3]') — older papers asked them, but they are removed from the rationalised NCERT book and are not in the UPMSP 2026-27 syllabus. Skip.रोले का प्रमेय तथा माध्यमान प्रमेय (जैसे '[1, 3] पर x³ − 5x² − 3x के लिए माध्यमान प्रमेय सत्यापित कीजिए') — पुराने प्रश्नपत्रों में पूछे गए, परन्तु संशोधित NCERT पुस्तक से हटा दिए गए हैं और UPMSP 2026-27 पाठ्यक्रम में नहीं हैं। छोड़ दीजिए।Rolle's theorem aur Mean Value Theorem (jaise '[1, 3] par x³ − 5x² − 3x ke liye MVT verify kijiye') — purane papers mein poochhe gaye, lekin rationalised NCERT book se hata diye gaye hain aur UPMSP 2026-27 syllabus mein nahi hain. Skip kijiye.
📖 Full chapter on the siteपूरा अध्याय साइट परPoora chapter site par📝 All Maths chaptersगणित के सभी अध्यायMaths ke saare chapters

Notes written by Simple Learning Hub from the current NCERT syllabus. Year counts come only from UP Board papers we actually read.नोट्स Simple Learning Hub ने वर्तमान NCERT पाठ्यक्रम से लिखे हैं। वर्षों की गिनती केवल उन यूपी बोर्ड पेपरों से है जो हमने सच में पढ़े।Notes Simple Learning Hub ne current NCERT syllabus se likhe hain. Saalon ki ginti sirf un UP Board papers se hai jo humne sach mein padhe.