{"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 Question 1. Question 2. <\/p>\n Question 3. Question 4. Question 5. Question 6. Question 7. <\/p>\n Question 8. Question 9. Question 10. Question 11. Question 12. <\/p>\n Question 13. Question 14. Question 15. Question 16. Question 17. <\/p>\n Question 18. Question 19. Question 20. Question 21. Question 22. Question 23. <\/p>\n Question 24. Question 25. Question 26. Question 27. Question 28. <\/p>\n Question 29. Question 30. Question 31. Question 32. Question 33. Question 34. <\/p>\n Question 35. Question 36. 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}]}}TS Inter 1st Year Maths 1B Differentiation Important Questions Long Answer Type<\/h2>\n
\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
\n
\n<\/p>\n
\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
\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<\/p>\n
\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
\n<\/p>\n
\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
\n<\/p>\n
\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
\n<\/p>\n
\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
\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
\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<\/p>\n
\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
\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<\/p>\n
\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
\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
\n<\/p>\n
\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
\n
\n<\/p>\nSome More Maths 1B Differentiation Important Questions Long Answer Type<\/h3>\n
\nFind the derivative of 20log(tan x)<\/sup>.
\nSolution:
\nLet y = 20log(tan x)<\/sup>
\nDifferentiating on both sides with respect to ‘x’.
\n<\/p>\n
\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<\/p>\n
\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
\n<\/p>\n
\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<\/p>\n
\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<\/p>\n
\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<\/p>\n
\nFind the derivative of sec x from the first principle.
\nSolution:
\nGiven, f(x) = sec x
\nNow, f(x + h) = sec (x + h)
\n
\n
\n<\/p>\n
\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
\n
\n<\/p>\n
\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
\n<\/p>\n
\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
\n<\/p>\n
\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
\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<\/p>\n
\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
\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<\/p>\n
\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<\/p>\n
\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
\n<\/p>\n
\nFind the derivative of the function f(x) = (x2<\/sup> – 3)(4x3<\/sup> + 1). [May ’15 (AP)]
\nSolution:
\n<\/p>\n
\nFind the derivative of tan-1<\/sup>(log x). [Mar. ’19 (AP); May ’15 (TS)]
\nSolution:
\n<\/p>\n
\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
\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<\/p>\n
\nIf ay4<\/sup> = (x + b)5<\/sup> then 5yy11<\/sup> = (y1<\/sup>)2<\/sup>. [Mar. ’17 (TS)]
\nSolution:
\n
\n<\/p>\n
\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
\nIf y = \\(\\frac{2 x+3}{4 x+5}\\), then find y”.
\nSolution:
\ny = \\(\\frac{2 x+3}{4 x+5}\\)
\n<\/p>\n
\nIf f(x) = log(tan ex<\/sup>), then find f'(x). [Mar. ’19 (TS)]
\nSolution:
\n<\/p>\n
\nEvaluate \\({Lim}_{x \\rightarrow 0} \\frac{\\log _e(1+5 x)}{x}\\). [Mar. ’19 (TS)]
\nSolution:
\n<\/p>\n
\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<\/p>\n","protected":false},"excerpt":{"rendered":"