{"id":26467,"date":"2024-10-21T12:18:51","date_gmt":"2024-10-21T06:48:51","guid":{"rendered":"https:\/\/samacheerkalvi.guide\/?p=26467"},"modified":"2024-10-22T09:48:42","modified_gmt":"2024-10-22T04:18:42","slug":"samacheer-kalvi-11th-maths-guide-chapter-10-ex-10-4","status":"publish","type":"post","link":"https:\/\/samacheerkalvi.guide\/samacheer-kalvi-11th-maths-guide-chapter-10-ex-10-4\/","title":{"rendered":"Samacheer Kalvi 11th Maths Guide Chapter 10 Differentiability and Methods of Differentiation Ex 10.4"},"content":{"rendered":"

Tamilnadu State Board New Syllabus\u00a0Samacheer Kalvi 11th Maths Guide<\/a> Pdf Chapter 10 Differentiability and Methods of Differentiation Ex 10.4 Text Book Back Questions and Answers, Notes.<\/p>\n

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.4<\/h2>\n

Find the derivatives of the following:<\/p>\n

Question 1.
\ny = xcos x<\/sup>
\nAnswer:
\ny = xcos x<\/sup>
\nTaking log on both sides
\nlog y = log xcos x<\/sup>
\nlog y = cos x log x
\nDifferentiating with respect to x
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 2.
\ny = xlog x<\/sup> + (log x)x<\/sup>
\nAnswer:
\ny = xlog x<\/sup> + (log x)x<\/sup>
\nLet u = xlog x<\/sup>, v = (log x)x<\/sup>
\nlog u = log xlog x<\/sup>
\nlog u = (log x) (log x)
\nlog u = (log x)2<\/sup>
\n\"Samacheer
\nv = (log x)x<\/sup>
\nlog v = log (log x)x<\/sup>
\nlog v = x log (log x)
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 3.
\n\\(\\sqrt{x y}\\) = e(x – y)<\/sup>
\nAnswer:
\n\"Samacheer<\/p>\n

Question 4.
\nxy<\/sup> = yx<\/sup>
\nAnswer:
\nxy<\/sup> = yx<\/sup>
\nTaking log on both sides
\nlog xy<\/sup> = log yx<\/sup>
\ny log x = x log y
\nDifferentiate with respect to x
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 5.
\n(cos x)log x<\/sup>
\nAnswer:
\ny = (cos x)log x<\/sup>
\nTaking log on both sides
\nlog y = log (cos x)log x<\/sup>
\nlog y = (log x) log (cos x)
\nDifferentiating with respect to x
\n\"Samacheer<\/p>\n

Question 6.
\n\\(\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1\\)
\nAnswer:
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 7.
\n\\(\\sqrt{x^{2}+y^{2}}=\\tan ^{-1}\\left(\\frac{y}{x}\\right)\\)
\nAnswer:
\n\"Samacheer
\n\"Samacheer
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 8.
\ntan (x + y) + tan (x – y) = x
\nAnswer:
\ntan (x + y) + tan (x – y) = x
\nDifferentiating with respect to x
\n\"Samacheer<\/p>\n

Question 9.
\nIf cos(xy) = x, show that
\n\\(\\frac{d y}{d x}=\\frac{-(1+y \\sin (x y))}{x \\sin x y}\\)
\nAnswer:
\ncos (xy) = x
\nDifferentiating with respect to x
\n\"Samacheer<\/p>\n

Question 10.
\n\\(\\tan ^{-1} \\sqrt{\\frac{1-\\cos x}{1+\\cos x}}\\)
\nAnswer:
\nLet y = \\(\\tan ^{-1} \\sqrt{\\frac{1-\\cos x}{1+\\cos x}}\\)
\n[1 – cos 2\u03b8 = 2 sin2<\/sup>\u03b8 and 1 + cos 2\u03b8 = 2 sin2<\/sup> \u03b8]
\n\"Samacheer
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 11.
\ntan-1<\/sup> = \\(\\left(\\frac{6 x}{1-9 x^{2}}\\right)\\)
\nAnswer:
\n\"Samacheer
\n\"Samacheer<\/p>\n

Question 12.
\n\\(\\cos \\left(2 \\tan ^{-1} \\sqrt{\\frac{1-x}{1+x}}\\right)\\)
\nAnswer:
\n\"Samacheer
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 13.
\nx = a cost<\/sup> ; y = a sin3<\/sup>t
\nAnswer:
\nx = a cost<\/sup> , y = a sin3<\/sup>t
\n\"Samacheer<\/p>\n

Question 14.
\nx = a (cos t + t sin t);
\ny = a (sin t – t cos t)
\nAnswer:
\nx = a (cos t + t sin t) , y = a (sin t – t cos t)
\n\\(\\frac{d x}{d t}\\) = a [- sin t + t cos t + sin t ]
\n\\(\\frac{d x}{d t}\\) = at cos t —— (1)
\ny = a (sin t – t cos t)
\n\\(\\frac{d x}{d t}\\) = a [cos t – (t \u00d7 – sin t + cos t \u00d7 1)]
\n\\(\\frac{d x}{d t}\\) = a[cos t + t sin t – cos t]
\n\\(\\frac{d x}{d t}\\) = at sin t —— (2)
\nFrom equations (1) and (2) we get
\n\"Samacheer<\/p>\n

Question 15.
\nx = \\(\\frac{1-t^{2}}{1+t^{2}}\\) , y = \\(\\frac{2 t}{1+t^{2}}\\)
\nAnswer:
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 16.
\ncos-1<\/sup>\\(\\left(\\frac{1-x^{2}}{1+x^{2}}\\right)\\)
\nAnswer:
\nLet y = cos-1<\/sup>\\(\\left(\\frac{1-x^{2}}{1+x^{2}}\\right)\\)
\nPut x = tan \u03b8
\n\"Samacheer
\ny = cos-1<\/sup> (cos 2\u03b8)
\ny = 2\u03b8
\ny = 2 tan-1<\/sup> x
\n\"Samacheer<\/p>\n

Question 17.
\nsin-1<\/sup> (3x – 4x3<\/sup>)
\nAnswer:
\nLet y = sin-1<\/sup> (3x – 4x3<\/sup>)
\nPut x = sin \u03b8
\ny = sin-1<\/sup> (3 sin \u03b8 – 4 sin3<\/sup> \u03b8)
\ny = sin-1<\/sup> (sin 3\u03b8)
\ny = 3\u03b8
\ny = 3 sin-1<\/sup> x
\n\\(\\frac{d y}{d x}=\\frac{3}{\\sqrt{1-x^{2}}}\\)<\/p>\n

Question 18.
\ntan-1<\/sup> \\(\\left(\\frac{\\cos x+\\sin x}{\\cos x-\\sin x}\\right)\\)
\nAnswer:
\nLet y = tan-1<\/sup> \\(\\left(\\frac{\\cos x+\\sin x}{\\cos x-\\sin x}\\right)\\)
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 19.
\nFind the derivative of sin x2<\/sup> with respect to x2<\/sup>.
\nAnswer:
\nLet u = sin x2<\/sup>
\n\\(\\frac{\\mathrm{d} \\mathrm{u}}{\\mathrm{d} x}\\) = cos (x2<\/sup>) \u00d7 2x
\n\\(\\frac{\\mathrm{d} \\mathrm{u}}{\\mathrm{d} x}\\) = 2x cos (x2<\/sup>)
\nLet v = x2<\/sup>
\n\"Samacheer<\/p>\n

Question 20.
\nFind the derivative of sin-1<\/sup>\\(\\left(\\frac{2 x}{1+x^{2}}\\right)\\) with respect to tan-1<\/sup> x.
\nAnswer:
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 21.
\nIf u = tan-1<\/sup>\\(\\frac{\\sqrt{1+x^{2}}-1}{x}\\) and v = tan-1<\/sup>x, find \\(\\frac{\\mathrm{du}}{\\mathrm{dv}}\\)
\nAnswer:
\n\"Samacheer
\n\"Samacheer
\n\"Samacheer
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 22.
\nFind the derivative with tan-1<\/sup>\\(\\left(\\frac{\\sin x}{1+\\cos x}\\right)\\) with respect to tan-1<\/sup>\\(\\left(\\frac{\\cos x}{1+\\sin x}\\right)\\)
\nAnswer:
\n\"Samacheer
\n\"Samacheer
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 23.
\nIf y = sin-1<\/sup>x then find y”.
\nAnswer:
\ny = sin-1<\/sup> x
\n\"Samacheer<\/p>\n

Question 24.
\nIf y = etan-1<\/sup><\/sup>x, show that (1 + x2<\/sup>) y” + (2x – 1) y’ = 0
\nAnswer:
\ny = etan-1x<\/sup>
\ny = etan-1x<\/sup> \\(\\left(\\frac{1}{1+x^{2}}\\right)\\)
\n\u21d2 y’ = \\(\\frac{y}{1+x^{2}}\\) \u21d2 y'(1 + x2<\/sup>) = y
\ndifferentiating w.r.to x
\ny’ (2x) + (1 + x2<\/sup>) (y”) = y’
\n(i.e.) (1 + x2<\/sup>) y” + y’ (2x) – y’ = 0
\n(i.e.) (1 + x2<\/sup>) y” + (2x – 1) y’ = 0<\/p>\n

Question 25.
\nIf y = \\(\\frac{\\sin ^{-1} x}{\\sqrt{1-x^{2}}}\\), show that (1 – x2<\/sup>)y2<\/sub> – 3xy1<\/sub> – y = 0.
\nAnswer:
\n\"Samacheer
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 26.
\nIf x = a (\u03b8 + sin \u03b8), y = a (1 – cos \u03b8) then prove that at \u03b8 = \\(\\frac{\\pi}{2}\\), y” = \\(\\frac{1}{a}\\)
\nAnswer:
\n\"Samacheer
\n\"Samacheer<\/p>\n

Question 27.
\nIf sin y = x sin (a + y), the prove that \\(\\frac{d y}{d x}=\\frac{\\sin ^{2}(a+y)}{\\sin a}\\), a \u2260 n\u03c0
\nAnswer:
\nGiven sin y = x sin (a + y) ——- (1)
\nDifferentiating with respect to x , we get
\ncos y \\(\\frac{\\mathrm{dy}}{\\mathrm{d} x}\\) = x cos (a + y) (0 + \\(\\frac{\\mathrm{dy}}{\\mathrm{d} x}\\)) + sin (a + y) . 1
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 28.
\nIf y = (cos-1<\/sup> x)2<\/sup>, prove that (1 – x2<\/sup>) \\(\\frac{\\mathrm{d}^{2} \\mathrm{y}}{\\mathrm{d} x^{2}}\\) – x \\(\\frac{\\mathrm{dy}}{\\mathrm{d} x}\\) – 2 = 0. Hence find y2<\/sub> when x = 0.
\nAnswer:
\n\"Samacheer
\n\"Samacheer<\/p>\n","protected":false},"excerpt":{"rendered":"

Tamilnadu State Board New Syllabus\u00a0Samacheer Kalvi 11th Maths Guide Pdf Chapter 10 Differentiability and Methods of Differentiation Ex 10.4 Text Book Back Questions and Answers, Notes. Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.4 Find the derivatives of the following: Question 1. y = xcos x Answer: y …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[6],"tags":[],"class_list":["post-26467","post","type-post","status-publish","format-standard","hentry","category-class-11"],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/26467"}],"collection":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/comments?post=26467"}],"version-history":[{"count":1,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/26467\/revisions"}],"predecessor-version":[{"id":41387,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/26467\/revisions\/41387"}],"wp:attachment":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/media?parent=26467"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/categories?post=26467"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/tags?post=26467"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}