{"id":5218,"date":"2024-01-04T16:54:21","date_gmt":"2024-01-04T11:24:21","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=5218"},"modified":"2024-01-06T09:54:21","modified_gmt":"2024-01-06T04:24:21","slug":"maths-1b-differentiation-important-questions-long-answer-type","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/maths-1b-differentiation-important-questions-long-answer-type\/","title":{"rendered":"TS Inter 1st Year Maths 1B Differentiation Important Questions Long Answer Type"},"content":{"rendered":"

Students must practice these Maths 1B Important Questions<\/a> TS Inter 1st Year Maths 1B Differentiation Important Questions Long Answer Type to help strengthen their preparations for exams.<\/p>\n

TS Inter 1st Year Maths 1B Differentiation Important Questions Long Answer Type<\/h2>\n

Question 1.
\nIf y = \\(\\tan ^{-1}\\left[\\frac{\\sqrt{1+x^2}+\\sqrt{1-x^2}}{\\sqrt{1+x^2}-\\sqrt{1-x^2}}\\right]\\), then find \\(\\frac{\\mathbf{d y}}{\\mathbf{d x}}\\). [Mar. ’18 (TS); Mar. ’16 (AP), ’12, ’10, ’09, ’04; May ’15 (AP & TS), ’12, ’97]
\nSolution:
\n\"TS
\n\"TS
\n\"TS<\/p>\n

Question 2.
\nIf y = xtan x<\/sup> + (sin x)cos x<\/sup>, then find \\(\\frac{\\mathbf{d y}}{\\mathbf{d x}}\\). [Mar. ’14, ’13 (Old), ’11, ’08, ’07; May ’13, ’06; Mar. ’18 (AP)]
\nSolution:
\nGiven y = xtan x<\/sup> + (sin x)cos x<\/sup>
\nDifferentiating on both sides with respect to ‘x’.
\nLet u = xtan x<\/sup>
\nv = sin xcos x<\/sup>
\ny = u + v
\n\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}=\\frac{\\mathrm{d}}{\\mathrm{dx}}(\\mathrm{u}+\\mathrm{v})\\)
\n\\(\\frac{d y}{d x}=\\frac{d u}{d x}+\\frac{d v}{d x}\\) ……(1)
\nNow, u = xtan x<\/sup>
\nTaking logarithms on both sides,
\nlog u = log xtan x<\/sup>
\nlog u = tan x log x
\nDifferentiating on both sides with respect to ‘x’.
\n\"TS
\nNow, v = (sin x)cos x<\/sup>
\nTaking logarithms on both sides,
\nlog v = log (sin x)cos x<\/sup>
\nlog v = cos x log(sin x)
\nDifferentiating on both sides with respect to ‘x’.
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 3.
\nIf x = \\(\\frac{3 a t}{1+t^3}\\), y = \\(\\frac{3 \\mathrm{at}^2}{1+\\mathrm{t}^3}\\), then find \\(\\frac{\\mathbf{d y}}{\\mathbf{d x}}\\). [B.P.]
\nSolution:
\nGiven that x = \\(\\frac{3 a t}{1+t^3}\\)
\nDifferentiating on both sides with respect to ‘t’.
\n\"TS
\n\"TS<\/p>\n

Question 4.
\nIf \\(\\sqrt{1-x^2}+\\sqrt{1-y^2}\\) = a(x – y) then show that \\(\\frac{d y}{d x}=\\sqrt{\\frac{1-y^2}{1-x^2}}\\). [Mar. ’17 (TS), ’08, ’05; May ’14, ’13 (Old), ’11, ’97]
\nSolution:
\nGiven that \\(\\sqrt{1-x^2}+\\sqrt{1-y^2}\\) = a(x – y)
\nPut x = sin \u03b1 \u21d2 \u03b1 = sin-1<\/sup>x
\ny = sin \u03b2 \u21d2 \u03b2 = sin-1<\/sup>y
\nNow, \\(\\sqrt{1-\\sin ^2 \\alpha}+\\sqrt{1-\\sin ^2 \\beta}\\) = a(sin \u03b1 – sin \u03b2)
\n\\(\\sqrt{\\cos ^2 \\alpha}+\\sqrt{\\cos ^2 \\beta}\\) = a(sin \u03b1 – sin \u03b2)
\ncos \u03b1 + cos \u03b2 = a(sin \u03b1 – sin \u03b2)
\n\"TS
\n\"TS<\/p>\n

Question 5.
\nIf y = \\(\\mathbf{x} \\sqrt{\\mathbf{a}^2+x^2}+a^2 \\log \\left(x+\\sqrt{a^2+x^2}\\right)\\) then show that \\(\\frac{d y}{d x}=2 \\sqrt{a^2+x^2}\\). [Mar. ’19, ’15 (AP). ’09, ’02; May ’08]
\nSolution:
\nGiven, that y = \\(\\mathbf{x} \\sqrt{\\mathbf{a}^2+x^2}+a^2 \\log \\left(x+\\sqrt{a^2+x^2}\\right)\\)
\nDifferentiating on both sides with respect to ‘x’.
\n\"TS
\n\"TS<\/p>\n

Question 6.
\nIf y = \\(\\tan ^{-1}\\left(\\frac{2 x}{1-x^2}\\right)+\\tan ^{-1}\\left(\\frac{3 x-x^3}{1-3 x^2}\\right)\\) – \\(\\tan ^{-1}\\left(\\frac{4 x-4 x^3}{1-6 x^2+x^4}\\right)\\) then show that \\(\\frac{\\mathbf{d y}}{\\mathbf{d x}}=\\frac{1}{1+x^2}\\). [May ’07]
\nSolution:
\nGiven that
\n\"TS
\ny = 2\u03b8 + 3\u03b8 – 4\u03b8 = \u03b8
\ny = tan-1<\/sup>x
\nDifferentiating on both sides with respect to x.
\n\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}=\\frac{\\mathrm{d}}{\\mathrm{dx}} \\tan ^{-1} \\mathrm{x}\\)
\n\u2234 \\(\\frac{d y}{d x}=\\frac{1}{1+x^2}\\)<\/p>\n

Question 7.
\nIf y = \\(\\frac{(1-2 x)^{2 \/ 3}(1+3 x)^{-3 \/ 4}}{(1-6 x)^{5 \/ 6}(1+7 x)^{-6 \/ 7}}\\), then find \\(\\frac{\\mathbf{d y}}{\\mathbf{d x}}\\). [May ’10]
\nSolution:
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 8.
\nIf y = (sin x)log x<\/sup> + xsin x<\/sup> then find \\(\\frac{\\mathbf{d y}}{\\mathbf{d x}}\\). [Mar. ’17 (AP), ’15 (TS), ’13]
\nSolution:
\nGiven that, let y = (sin x)log x<\/sup> + xsin x<\/sup>
\nLet, u = (sin x)log x<\/sup> , v = xsin x<\/sup> then y = u + v
\nDifferentiating on both sides with respect to x.
\n\\(\\frac{d y}{d x}=\\frac{d u}{d x}+\\frac{d v}{d x}\\) …….(1)
\nNow, u = (sin x)log x<\/sup>
\nTaking logarithms on both sides
\nlog u = log (sin x)log x<\/sup>
\nlog u = log x . log (sin x)
\nDifferentiating on both sides with respect to ‘x’.
\n\"TS
\nNow, v = xsin x<\/sup>
\nTaking logarithms on both sides
\nlog v = log xsin x<\/sup>
\nlog v = sin x log x
\nDifferentiating on both sides with respect to ‘x’.
\n\"TS<\/p>\n

Question 9.
\nIf xy<\/sup> + yx<\/sup> = ab<\/sup> then show that \\(\\frac{d y}{d x}=-\\left[\\frac{y x^{y-1}+y^x \\log y}{x^y \\log x+x y^{x-1}}\\right]\\). [Mar. ’03; Mar. ’16 (TS)]
\nSolution:
\nGiven that, xy<\/sup> + yx<\/sup> = ab<\/sup>
\nLet, xy<\/sup> = u, yx<\/sup> = v then, u + v = ab<\/sup>
\nDifferentiating on both sides with respect to x.
\n\\(\\frac{\\mathrm{du}}{\\mathrm{dx}}+\\frac{\\mathrm{dv}}{\\mathrm{dx}}=0\\) ……(1)
\nNow, u = xy<\/sup>
\nTaking logarithms on both sides
\nlog u = log xy<\/sup>
\nlog u = y log x
\nDifferentiating on both sides with respect to ‘x’.
\n\"TS
\nNow, v = yx<\/sup>
\nTaking logarithms on both sides
\nlog v = log yx<\/sup>
\nlog v = x log y
\nDifferentiating on both sides with respect to ‘x’.
\n\"TS
\n\"TS<\/p>\n

Question 10.
\nIf f(x) = \\(\\sin ^{-1} \\sqrt{\\frac{x-\\beta}{\\alpha-\\beta}}\\) and g(x) = \\(\\tan ^{-1} \\sqrt{\\frac{x-\\beta}{\\alpha-x}}\\) then show that f'(x) = g'(x), (\u03b2 < x < \u03b1).
\nSolution:
\n\"TS
\n\"TS
\n\"TS<\/p>\n

Some More Maths 1B Differentiation Important Questions Long Answer Type<\/h3>\n

Question 11.
\nFind the derivative of 20log(tan x)<\/sup>.
\nSolution:
\nLet y = 20log(tan x)<\/sup>
\nDifferentiating on both sides with respect to ‘x’.
\n\"TS<\/p>\n

Question 12.
\nIf f(x) = e2x<\/sup> log x (x > 0), then find f'(x).
\nSolution:
\nf(x) = e2x<\/sup> (log x)
\nLet y = e2x<\/sup> (log x)
\nDifferentiating with respect to ‘x’ on both sides,
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 13.
\nIf y = \\(\\frac{\\mathbf{a}-\\mathbf{x}}{\\mathbf{a}+\\mathbf{x}}\\), find \\(\\frac{\\mathbf{d y}}{\\mathbf{d x}}\\).
\nSolution:
\nGiven y = \\(\\frac{\\mathbf{a}-\\mathbf{x}}{\\mathbf{a}+\\mathbf{x}}\\)
\nDifferentiating with respect to x on both sides.
\n\"TS
\n\"TS<\/p>\n

Question 14.
\nIf y = sin-1<\/sup>(cos x) then find \\(\\frac{d y}{d x}\\).
\nSolution:
\nLet y = sin-1<\/sup>(cos x)
\nDifferentiating on both sides with respect to ‘x’.
\n\"TS<\/p>\n

Question 15.
\nIf x4<\/sup> + y4<\/sup> – a2<\/sup>xy = 0, find \\(\\frac{\\mathbf{d y}}{\\mathbf{d x}}\\).
\nSolution:
\nGiven that x4<\/sup> + y4<\/sup> – a2<\/sup>xy = 0
\nDifferentiating on both sides with respect to ‘x’.
\n\"TS<\/p>\n

Question 16.
\nFind the derivative of cos-1<\/sup>(4x3<\/sup> – 3x) with respect to ‘x’.
\nSolution:
\nLet y = cos-1<\/sup>(4x3<\/sup> – 3x)
\nPut x = cos \u03b8
\n\u21d2 \u03b8 = cos-1<\/sup>x
\nNow, y = cos-1<\/sup>(4 cos3<\/sup>\u03b8 – 3 cos \u03b8)
\n= cos-1<\/sup>(cos 3\u03b8)
\n= 3\u03b8
\ny = 3 cos-1<\/sup>x
\nDifferentiating on both sides with respect to ‘x’.
\n\"TS<\/p>\n

Question 17.
\nFind the derivative of sec x from the first principle.
\nSolution:
\nGiven, f(x) = sec x
\nNow, f(x + h) = sec (x + h)
\n\"TS
\n\"TS
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 18.
\nFind the derivative of \\(\\sin ^{-1}\\left(\\frac{b+a \\sin x}{a+b \\sin x}\\right)\\).
\nSolution:
\nLet y = \\(\\sin ^{-1}\\left(\\frac{b+a \\sin x}{a+b \\sin x}\\right)\\)
\nDifferentiating on both sides with respect to ‘x’.
\n\"TS
\n\"TS
\n\"TS<\/p>\n

Question 19.
\nFind the derivative of \\(\\tan ^{-1}\\left(\\frac{\\cos x}{1+\\cos x}\\right)\\)
\nSolution:
\nLet y = \\(\\tan ^{-1}\\left(\\frac{\\cos x}{1+\\cos x}\\right)\\)
\nDifferentiating on both sides with respect to ‘x’.
\n\"TS
\n\"TS<\/p>\n

Question 20.
\nFind the derivative of \\(\\tan ^{-1}\\left(\\frac{\\sqrt{1+x^2}-1}{x}\\right)\\) with respect to tan-1<\/sup>x. [May ’09]
\nSolution:
\n\"TS
\n\"TS<\/p>\n

Question 21.
\nIf f(x) = x ex<\/sup> sin x, then find f'(x).
\nSolution:
\nf(x) = x . ex<\/sup> . sin x
\nLet y = x . ex<\/sup> . sin x
\n\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = \\(\\frac{d}{d x}\\) (x . ex<\/sup> . sin x)
\n= x . ex<\/sup> . \\(\\frac{d}{d x}\\) (sin x) + x . sin x . \\(\\frac{d}{d x}\\) (ex<\/sup>) + sin x . ex<\/sup> . \\(\\frac{d}{d x}\\) (x)
\n= x . ex<\/sup> . cos x + x . sin x . ex<\/sup> + sin x . ex<\/sup> (1)
\n= ex<\/sup> (x cos x + x sin x + sin x)
\n\u2234 f'(x) = ex<\/sup> (x cos x + x sin x + sin x)<\/p>\n

Question 22.
\nIf f(x) = sin(log x), (x > 0), then find f'(x). [Mar. ’18 (AP)]
\nSolution:
\nf(x) = sin(log x)
\nLet y = sin(log x)
\n\"TS<\/p>\n

Question 23.
\nIf f(x) = (x3<\/sup> + 6x2<\/sup> + 12x – 13)100<\/sup>, then find f'(x).
\nSolution:
\nGiven that, f(x) = (x3<\/sup> + 6x2<\/sup> + 12x – 13)100<\/sup>
\nLet y = (x3<\/sup> + 6x2<\/sup> + 12x – 13)100<\/sup>
\nDifferentiating with respect to x on both sides.
\n\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = \\(\\frac{d}{d x}\\) (x3<\/sup> + 6x2<\/sup> + 12x – 13)100<\/sup>
\n= 100(x3<\/sup> + 6x2<\/sup> + 12x – 13)100-1<\/sup> \\(\\frac{d}{d x}\\)(x3<\/sup> + 6x2<\/sup> + 12x – 13)
\n= 100(x3<\/sup> + 6x2<\/sup> + 12x – 13)99<\/sup> (3x2<\/sup> + 6(2x) + 12(1) – 0)
\n= 100(x3<\/sup> + 6x2<\/sup> + 12x – 13)99<\/sup> (3x2<\/sup> + 12x + 12)
\n= 100(x3<\/sup> + 6x2<\/sup> + 12x – 13)99<\/sup> 3(x2<\/sup> + 4x + 4)
\n= 300(x + 2)2<\/sup> (x3<\/sup> + 6x2<\/sup> + 12x – 13)99<\/sup>
\n\u2234 f'(x) = 300(x + 2)2<\/sup> . (x3<\/sup> + 6x2<\/sup> + 12x – 13)99<\/sup><\/p>\n

\"TS<\/p>\n

Question 24.
\nFind the derivative of (ax + b)n<\/sup> (cx + d)m<\/sup>.
\nSolution:
\nGiven, f(x) = (ax + b)n<\/sup> (cx + d)m<\/sup>
\nLet y = (ax + b)n<\/sup> (cx + d)m<\/sup>
\nDifferentiating with respect to ‘x’ on both sides.
\n\"TS<\/p>\n

Question 25.
\nFind the derivative of \\(\\frac{\\mathbf{p} \\mathbf{x}^2+\\mathbf{q x}+\\mathbf{r}}{\\mathbf{a x}+\\mathbf{b}}\\).
\nSolution:
\nGiven, f(x) = \\(\\frac{\\mathbf{p} \\mathbf{x}^2+\\mathbf{q x}+\\mathbf{r}}{\\mathbf{a x}+\\mathbf{b}}\\)
\nLet y = \\(\\frac{\\mathbf{p} \\mathbf{x}^2+\\mathbf{q x}+\\mathbf{r}}{\\mathbf{a x}+\\mathbf{b}}\\)
\nDifferentiating with respect to ‘x’ on both sides.
\n\"TS<\/p>\n

Question 26.
\nFind the derivative of \\(\\log _7(\\log x)\\).
\nSolution:
\nGiven that, f(x) = \\(\\log _7(\\log x)\\)
\nLet y = \\(\\log _7(\\log x)\\)
\nDifferentiating with respect to ‘x’ on both sides.
\n\"TS
\n\"TS<\/p>\n

Question 27.
\nFind the derivative of the function f(x) = (x2<\/sup> – 3)(4x3<\/sup> + 1). [May ’15 (AP)]
\nSolution:
\n\"TS<\/p>\n

Question 28.
\nFind the derivative of tan-1<\/sup>(log x). [Mar. ’19 (AP); May ’15 (TS)]
\nSolution:
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 29.
\nIf f(x) = 2x3<\/sup> + 3x – 5, then prove that f'(0) + 3 . f'(-1) = 0. [Mar. ’16 (AP)]
\nSolution:
\nGiven f(x) = 2x2<\/sup> + 3x – 5
\nNow f'(x) = 2(2x) + 3(1) – 0 = 4x + 3
\nf'(0) = 4(0) + 3 = 3
\nf(-1) = 4(-1) + 3 = -4 + 3 = -1
\nLHS = f'(0) + 3. f'(-1)
\n= 3 + 3(-1)
\n= 3 – 3
\n= 0
\n\u2234 f'(0) + 3 . f'(-1) = 0<\/p>\n

Question 30.
\nIf 2x2<\/sup> – 3xy + y2<\/sup> + x + 2y – 8 = 0, then find \\(\\frac{d \\mathbf{y}}{\\mathbf{d x}}\\). [Mar. ’16 (TS)]
\nSolution:
\nGiven 2x2<\/sup> – 3xy + y2<\/sup> + 2y – 8 = 0
\ndifferentiating on both sides with respect to ‘x’.
\n\"TS<\/p>\n

Question 31.
\nIf ay4<\/sup> = (x + b)5<\/sup> then 5yy11<\/sup> = (y1<\/sup>)2<\/sup>. [Mar. ’17 (TS)]
\nSolution:
\n\"TS
\n\"TS<\/p>\n

Question 32.
\nIf y = x4<\/sup> + tan x then find y11<\/sup>. [Mar. ’18 (AP)]
\nSolution:
\nGiven that y = x4<\/sup> + tan x
\ny1<\/sup> = \\(\\frac{\\mathrm{d}}{\\mathrm{dx}}\\)(x4<\/sup> + tan x) = 4x3<\/sup> + sec2<\/sup>x
\ny11<\/sup> = \\(\\frac{\\mathrm{d}}{\\mathrm{dx}}\\)(4x3<\/sup> + sec2<\/sup>x)
\n= 4(3x2<\/sup>) + 2 sec x \\(\\frac{\\mathrm{d}}{\\mathrm{dx}}\\)(sec x)
\n= 12x2<\/sup> + 2 sec x (sec x tan x)
\n\u2234 y11<\/sup> = 12x2<\/sup> + 2 sec2<\/sup>x tan x<\/p>\n

Question 33.
\nIf y = \\(\\frac{2 x+3}{4 x+5}\\), then find y”.
\nSolution:
\ny = \\(\\frac{2 x+3}{4 x+5}\\)
\n\"TS<\/p>\n

Question 34.
\nIf f(x) = log(tan ex<\/sup>), then find f'(x). [Mar. ’19 (TS)]
\nSolution:
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 35.
\nEvaluate \\({Lim}_{x \\rightarrow 0} \\frac{\\log _e(1+5 x)}{x}\\). [Mar. ’19 (TS)]
\nSolution:
\n\"TS<\/p>\n

Question 36.
\nIf xlog y<\/sup> = log x, then show that \\(\\frac{d y}{d x}=\\frac{y}{x}\\left(\\frac{1-\\log x \\log y}{(\\log x)^2}\\right)\\). [Mar. ’19 (TS)]
\nSolution:
\n\"TS<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice these Maths 1B Important Questions TS Inter 1st Year Maths 1B Differentiation Important Questions Long Answer Type to help strengthen their preparations for exams. TS Inter 1st Year Maths 1B Differentiation Important Questions Long Answer Type Question 1. If y = , then find . [Mar. ’18 (TS); Mar. ’16 (AP), ’12, … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/5218"}],"collection":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=5218"}],"version-history":[{"count":4,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/5218\/revisions"}],"predecessor-version":[{"id":5714,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/5218\/revisions\/5714"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=5218"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=5218"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=5218"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}