{"id":20104,"date":"2020-01-17T15:28:30","date_gmt":"2020-01-17T09:58:30","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=20104"},"modified":"2021-07-02T15:28:51","modified_gmt":"2021-07-02T09:58:51","slug":"2nd-puc-basic-maths-question-bank-chapter-18-ex-18-6","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-basic-maths-question-bank-chapter-18-ex-18-6\/","title":{"rendered":"2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.6"},"content":{"rendered":"

Students can Download Basic Maths Exercise 18.6 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 18.6<\/h2>\n

Part – A<\/span><\/p>\n

2nd PUC Basic Maths Differential Calculus Ex 18.6 Two or Three Marks Questions and Answers<\/strong><\/p>\n

Question 1.
\nFind \\(\\frac{d y}{d x}\\) if
\n(a) x = \\(a\\left(t-\\frac{1}{t}\\right), y=a\\left(t+\\frac{1}{t}\\right) 1\\)
\n(b) x = e2t<\/sup> , y = log(2t + 1)
\n(c) x = log(1 + t), y = \\(\\frac{1}{1+t}\\)
\n(d) x = log t, y = \\(\\frac{1}{t}\\)
\n(e) x = 4t , y = \\(\\frac{4}{t}\\)
\n(f) x = a sec \u03b8, y = b tan \u03b8.
\n(g) x = a(\u03b8 – sin \u03b8), y = a(1 – cos \u03b8)
\n(h) x = a cos(log t), y = a log(cos t)
\nAnswer:
\n(a) Given x = \\(a\\left(t-\\frac{1}{t}\\right), y=a\\left(t+\\frac{1}{t}\\right) 1\\)
\nDifferentiate both w.r.t
\n\"2nd<\/p>\n

(b) Given x = e2t<\/sup> , y = log(2t + 1)
\nDifferentiate both w.r.t we get
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

(c) Given x = log (1+t), y = \\(\\frac{1}{1+t}\\)
\nDifferentiate both w.r.t x t, we get
\n\"2nd<\/p>\n

(d) Given x = log t, y = \\(\\frac{1}{t}\\)
\nDifferentiate both w.r.t t, we get
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

(e) Given x = 4t y = \\(\\frac{4}{t}\\)
\nDifferentiate both w.r.t t, we get
\n\"2nd<\/p>\n

(f) Given a = sec\u03b8, y = b tan \u03b8
\nDifferentiate both w.r.t. \u03b8 we get
\n\"2nd<\/p>\n

(g) Given x = a (\u03b8 – sin \u03b8), y = a(1 – cos\u03b8)
\nDifferentiate both w.r.t \u03b8
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

(h) Given x = a cos(logt) y = a log(cos t)
\nDifferentiate both w.r.t t
\n\"2nd<\/p>\n

Question 2.
\nDifferentiate tan2<\/sup> w.r.t cos2<\/sup>x.
\nAnswer:
\nLet u = tan2<\/sup>x, v = cos2<\/sup>x
\nDifferentiate both w.r.t x
\n\\(\\frac{d u}{d x}\\) = 2tan x. sec2<\/sup>x,
\n\\(\\frac{d v}{d x}\\) = 2 cosx(-sinx)
\n\u2234 \\(\\frac{d u}{d v}\\) = \\(\\frac{\\frac{d u}{d x}}{\\frac{d v}{d x}}=\\frac{2 \\tan x \\cdot \\sec ^{2} x}{2 \\sin x \\cdot \\cos x}\\)
\n= \\(\\frac{1}{\\sec ^{4} x}\\)<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nDifferentiate sin2<\/sup>x w.r.t. x2<\/sup>
\nAnswer:
\nLet u = sin2<\/sup>x, v = x2<\/sup>
\nDifferentiate both w.r.t. x
\n\\(\\frac{d u}{d x}\\) = 2 sin x. cos x,
\n\\(\\frac{d v}{d x}\\) = 2x
\n\u2234 \\(\\frac{d u}{d v}\\) = \\(\\frac{\\frac{d u}{d x}}{\\frac{d v}{d x}}=\\frac{2 \\sin x \\cos x}{2 x}\\)
\n= \\(\\frac{\\sin x \\cos x}{x}\\)<\/p>\n

Question 4.
\nDifferentiate tan \\(\\sqrt{x}\\) w.r.t \\(\\sqrt{x}\\)
\nAnswer:
\nLet u = tan \\(\\sqrt{x}\\) v = \\(\\sqrt{x}\\)
\nDifferentiate both w.r.t x
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nDifferentiate log x w.r.t \\(\\frac { 1 }{ x }\\)
\nAnswer:
\nLet u = log x, v = \\(\\frac { 1 }{ x }\\)
\n\"2nd<\/p>\n

Question 6.
\nDifferentiate log sin x w.r.t \\(\\sqrt{\\cos x}\\)
\nAnswer:
\nLet u = log (sin x) v = \\(\\sqrt{\\cos x}\\)
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 7.
\nIf x = elog cos 4\u03b8<\/sup> , y = elog sin 4\u03b8<\/sup> show that \\(\\frac{d y}{d x}=\\frac{-x}{y}\\)
\nAnswer:
\nGiven x = cos 4\u03b8, y = sin 4\u03b8
\n\\(\\frac{d x}{d \\theta}\\) = -4sin 4\u03b8
\n\\(\\frac{d y}{d \\theta}\\) = + 4cos 4\u03b8
\n\"2nd<\/p>\n

Question 8.
\nIf x = a cos4<\/sup>\u03b8, y = asin 4\u03b8<\/sup> show that \\(\\frac{d y}{d x}\\) = -tan2<\/sup>\u03b8
\nAnswer:
\nGiven x = a cos4<\/sup>\u03b8, y = a sin 4\u03b8
\n\\(\\frac{d x}{d \\theta}\\) = a(4cos3<\/sup>\u03b8( – sin\u03b8),) \\(\\frac{d y}{d \\theta}\\) = 4a sin3<\/sup>\u03b8 cos \u03b8
\n\u2234 \\(\\frac{d y}{d x}=\\frac{4 a \\sin ^{3} \\theta \\cos \\theta}{-4 a \\cos ^{3} \\theta \\sin \\theta}=-\\tan ^{2} \\theta\\)<\/p>\n

Question 9.
\nIf x = et<\/sup>(cos t + sin t), y = et<\/sup>(cos t – sin t). show that \\(\\frac{d y}{d x}\\) = -tan t.
\nAnswer:
\nGiven x = et<\/sup>(cos t + sin t), y = et<\/sup>(cos t – sin t)
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 10.
\nIf x = a log sec \u03b8, y = a(tan\u03b8 – 1) show that \\(\\frac{d y}{d x}\\) = 2 cosec2\u03b8
\nAnswer:
\nGiven x = a log sec \u03b8, y = a(tan\u03b8 – 1)
\n\"2nd<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Basic Maths Exercise 18.6 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 18.6 Part – A 2nd PUC …<\/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\/20104"}],"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=20104"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/20104\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=20104"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=20104"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=20104"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}