{"id":19792,"date":"2020-01-14T15:27:08","date_gmt":"2020-01-14T09:57:08","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=19792"},"modified":"2021-07-02T15:28:59","modified_gmt":"2021-07-02T09:58:59","slug":"2nd-puc-basic-maths-question-bank-chapter-18-ex-18-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-basic-maths-question-bank-chapter-18-ex-18-2\/","title":{"rendered":"2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.2"},"content":{"rendered":"

Students can Download Basic Maths Exercise 18.2 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.2<\/h2>\n

Part-A<\/span><\/p>\n

2nd PUC Basic Maths Differential Calculus Ex 18.2 One or Two Marks Questions and Answers<\/strong><\/p>\n

Question 1.
\n(a2<\/sup> – x2<\/sup>)10<\/sup>
\nAnswer:
\nLet y(a2<\/sup> – x2<\/sup>)10<\/sup>
\n\\(\\frac{d y}{d x}\\) = 10(a2<\/sup> – x2<\/sup>)10 – 1<\/sup>. \\(\\frac{d y}{d x}\\)(a2<\/sup> – x2<\/sup>)
\n= 10(a2<\/sup> – x2<\/sup>)9<\/sup> (-2x) = -20x (a2<\/sup> – x2<\/sup>)9<\/sup><\/p>\n

Question 2.
\nlog[log(log x)]
\nAnswer:
\nLet y = log x (log log(x))
\n\\(\\frac{d y}{d x}=\\frac{1}{\\log (\\log x)} \\cdot \\frac{1}{\\log x} \\cdot \\frac{1}{x}\\)<\/p>\n

Question 3.
\ncos x3<\/sup>
\nAnswer:
\nLet y = cosx3<\/sup>
\n\\(\\frac{d y}{d x}\\) = -sinx3<\/sup> .3x2<\/sup><\/p>\n

\"KSEEB<\/p>\n

Question 4.
\nsin3<\/sup>\\(\\sqrt{x}\\)
\nAnswer:
\nLet y = sin3<\/sup>(\\(\\sqrt{x}\\)) = (sin \\(\\sqrt{x}\\)))3<\/sup>
\n\\(\\frac{d y}{d x}\\) = 3.sin2<\/sup> \\(\\sqrt{x}\\) . cos \\(\\sqrt{x}\\) . \\(\\frac{1}{2 \\sqrt{x}}\\)<\/p>\n

Question 5.
\n[log(cos x)]2<\/sup>
\nAnswer:
\nLet y = [log(cos x)]2<\/sup>
\n\\(\\frac{d y}{d x}\\) = 2log(cos x). \\(\\frac{1}{\\cos x}\\)(-sin x)
\n= -2 tan x log(cos x)<\/p>\n

Question 6.
\n\\(\\sec \\left(x+\\frac{1}{x}\\right)\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 7.
\n\\(7^{\\sin \\sqrt{x}}\\)
\nAnswer:
\nLet y = \\(7^{\\sin \\sqrt{x}}\\)
\n\\(\\frac{d y}{d x}\\) = \\(7^{\\sin \\sqrt{x}}\\) . log 7 . cos \\(\\sqrt{x}\\) . \\(\\frac{1}{2 \\sqrt{x}}\\)<\/p>\n

Question 8.
\n\\(\\sqrt{\\cot \\sqrt{x}}\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 9.
\nlog(sin \\(\\sqrt{x}\\))
\nAnswer:
\n\"2nd<\/p>\n

Question 10.
\nlog[log (tan x)]
\nAnswer:
\nLet y = log(log (tan x))
\n\\(\\frac{d y}{d x}\\) = \\(\\frac{1}{\\log (\\tan x)} \\cdot \\frac{1}{\\tan x} \\cdot \\sec ^{2} x\\)<\/p>\n

\"KSEEB<\/p>\n

Question 11.
\ncos 3x . sin 5x
\nAnswer:
\nLet y = cos 3x . sin 5x (Trans using formula)
\nsy = 2[sin 8x – sin(-2x)] = 2 [sin 8x] + 2sin 2x
\n\\(\\frac{d y}{d x}\\) = 16 cos 8x + 4 cos 2x
\nOR
\nLet y = cos 3x . sin 5x
\n\\(\\frac{d y}{d x}\\) = cos 3x(5 cos 5x) + sin 5x (-3 sin 3x) = 5 cos 3x sin5x – 3 sin 5x.sin 3x.<\/p>\n

Question 12.
\nsin x . sin 2x
\nAnswer:
\nLet y = sin x . sin 2x
\n\\(\\frac{d y}{d x}\\) = sin x(2 cos2x) + sin 2x cos x = 2 sin x cos 2x + cos x sin 2x<\/p>\n

Question 13.
\neloge<\/sub>(x + \\(\\sqrt{x^{2}+a^{2}}\\)).
\nAnswer:
\n\"2nd<\/p>\n

Question 14.
\ne2x<\/sup> . sin 3x.
\nAnswer:
\nLet y = e2x<\/sup> . sin 3x
\n\\(\\frac{d y}{d x}\\) = e2x<\/sup>(3 cos 3x) + sin 3x(2e2x<\/sup>)<\/p>\n

Question 15.
\ncos5<\/sup>x . cos(x5<\/sup>).
\nAnswer:
\nLet y = cos5<\/sup>x . cos(x5<\/sup>)
\n\\(\\frac{d y}{d x}\\) = cos5<\/sup>x(-sin(x5<\/sup>)5x4<\/sup>) + cos(x5<\/sup>).5cos4<\/sup>x.(-sin x)
\n= – cos5<\/sup>x sin(x5<\/sup>) 5x4<\/sup> – 5 cos(x5<\/sup>) . cos4<\/sup>x . sin x<\/p>\n

Question 16.
\n3x2<\/sup><\/sup> .log x.
\nAnswer:
\nLet y = 3x2<\/sup><\/sup> .log x.
\n\\(\\frac{d y}{d x}\\) = 3x2<\/sup><\/sup> . \\(\\frac { 1 }{ x }\\) + logx . 3x2<\/sup><\/sup> . loge<\/sub> 3 . 2x .<\/p>\n

Question 17.
\n\\(\\frac{x}{\\sqrt{x^{2}-1}}\\)
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 18.
\n\\(\\frac{x}{\\sqrt{2 x-1}}\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 19.
\n\\(\\frac{\\mathrm{e}^{\\sin \\mathrm{x}}}{\\sqrt{\\log \\mathrm{x}}}\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 20.
\n\\(\\log \\left(\\frac{1+\\sin x}{1-\\sin x}\\right)\\)
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Part-B<\/span><\/p>\n

2nd PUC Basic Maths Differential Calculus Ex 18.2 Three Marks Questions and Answers<\/strong><\/p>\n

Question 1.
\nIf y = \\(\\left(\\frac{\\cos x+\\sin x}{\\cos x-\\sin x}\\right)\\) , show that \\(\\frac{d y}{d x}\\) = sec2<\/sup> \\(\\left(x+\\frac{\\pi}{4}\\right)\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 2.
\nIf y = log \\(\\left[\\frac{1-\\cos x}{1+\\cos x}\\right]\\) , prove that \\(\\frac{d y}{d x}\\) = 2 cosec x.
\nAnswer:
\n\"2nd<\/p>\n

Question 3.
\nDifferentiate e2x<\/sup> w.r.t x from first principles
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\nDifferentiate sin 2x w.r.t x from first principles.
\nAnswer:
\n\"2nd<\/p>\n

Question 5.
\nDifferentiate tan ax w.r.t x froom the principles.
\nAnswers:
\n\"2nd<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Basic Maths Exercise 18.2 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.2 Part-A 2nd PUC Basic Maths …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[82],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/19792"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=19792"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/19792\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=19792"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=19792"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=19792"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}