{"id":35115,"date":"2022-08-25T10:40:17","date_gmt":"2022-08-25T05:10:17","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=35115"},"modified":"2022-08-25T10:40:17","modified_gmt":"2022-08-25T05:10:17","slug":"inter-1st-year-maths-1b-differentiation-solutions-ex-9d","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/inter-1st-year-maths-1b-differentiation-solutions-ex-9d\/","title":{"rendered":"Inter 1st Year Maths 1B Differentiation Solutions Ex 9(d)"},"content":{"rendered":"

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions<\/a> Inter 1st Year Maths 1B Differentiation Solutions Exercise 9(d) will help students to clear their doubts quickly.<\/p>\n

Intermediate 1st Year Maths 1B Differentiation Solutions Exercise 9(d)<\/h2>\n

I.<\/span><\/p>\n

Question 1.
\nIf y = \\(\\frac{2x+3}{4x+5}\\) then find y”
\nSolution:
\ny = \\(\\frac{2x+3}{4x+5}\\)
\nDifferentiating w.r.to x
\n\"Inter<\/p>\n

Question 2.
\ny = aenx<\/sup> + be-nx<\/sup> nx then prove that y” = n\u00b2y
\nSolution:
\ny = aenx<\/sup> + be-nx<\/sup>
\ny1<\/sub> = na enx<\/sup> – nb e-nx<\/sup>
\ny2 = n\u00b2 . aenx<\/sup> + n\u00b2 be-nx<\/sup>
\ny”= n\u00b2 (aenx<\/sup> + b.e-nx<\/sup>)
\n= n\u00b2y<\/sub><\/p>\n

II.<\/span><\/p>\n

Question 1.
\nFind the second order derivatives of the following functions f(x)
\ni) cos\u00b3 x
\nSolution:
\ny = cos\u00b3 x = \\(\\frac{1}{4}\\) [cos 3x + 3 cos x]
\ny’ = \\(\\frac{1}{4}\\) [- 3 sin 3x – 3 sin x]
\ny” = \\(\\frac{1}{4}\\) (- 9 cos 3x – 3 cos x)
\n= – \\(\\frac{1}{4}\\) (3 cos x + 9 cos 3x)
\n= – \\(\\frac{3}{4}\\) (cos x + 3 cos 3x)<\/p>\n

\"Inter<\/p>\n

ii) y = sin4<\/sup> x
\nSolution:
\ny = sin4<\/sup> x = (sin\u00b2x)\u00b2 = (\\(\\frac{(1-\\cos 2 x)^{2}}{2}\\))
\n= \\(\\frac{1}{4}\\) [1 – 2 cos 2x + cos\u00b2 2x]
\n= \\(\\frac{1}{4}\\) [1 – 2cos 2x + \\(\\frac{1+\\cos 4 x}{2}\\)]
\n= \\(\\frac{1}{8}\\) [2 – 4 cos 2x + 1 + cos 4x]
\n= \\(\\frac{1}{8}\\)(3 – 4 cos 2x + cos 4x)
\ny’ = \\(\\frac{1}{8}\\) (8 sin 2x – 4 sin 4x)
\ny” = \\(\\frac{1}{8}\\) (16 cos 2x-16 cos 4x)
\n= 2 (cos 2x – cos 4x)<\/p>\n

iii) log (4x\u00b2 – 9)
\nSolution:
\ny = log (4x\u00b2 – 9)
\n= log (2x – 3) (2x + 3)
\n= log (2x – 3) + log (2x + 3)
\n\"Inter<\/p>\n

iv) y = e-2x<\/sup> sin\u00b3 x
\nSolution:
\ny = e-2x<\/sup>. sin\u00b3 x
\ny’ = e-2x<\/sup> (3 sin\u00b2 x. cos x) + sin3 x (e-2x<\/sup>) (-2)
\n= e-2x<\/sup> [3 sin\u00b2 x. cos x – 2 sin\u00b3 x)
\n\\(\\frac{d^{2} y}{d x^{2}}\\) = e-2x<\/sup> [3 sin\u00b2 x (- sin x) + 3 cos x (2 sin x) cos x – 6 sin\u00b2 x cos x) – 2. e-2x<\/sup>
\n[3 sin\u00b2 x. cos x – 2 sin\u00b3 x]
\n= e-2x<\/sup> [6 sin x. cos\u00b2 x – 6 sin\u00b2 x. cos x – 3 sin\u00b3 x. – 6 sin\u00b2 x. cos x + 4 sin\u00b3 x)
\n= e-2x<\/sup> [sin\u00b3 x – 12 sin\u00b2 x. cos x + 6 sin x. cos\u00b2 x)<\/p>\n

v) ex<\/sup> sin x cos 2x
\nSolution:
\ny = ex<\/sup>. sin x. cos 2x = \\(\\frac{e^{x}}{2}\\) (2 cos 2x. sin x)
\n= \\(\\frac{e^{x}}{2}\\) (sin 3x – sin x)
\ny’ = \\(\\frac{1}{2}\\) [ex<\/sup>(3 cos 3x – cos x) + ex<\/sup> (sin 3x – sin x)
\ny”= \\(\\frac{1}{2}\\) [ex (- 9 sin 3x + sin x) + ex<\/sup> (3 cos 3x – cos x) + ex<\/sup> (3 cos 3x – cos x) + ex<\/sup> (sin 3x – sin x)]
\n= \\(\\frac{e^{x}}{2}\\) [- 9 sin 3x + sin x + 3 cos 3x – cos x + 3 cos 3x cos x + sin 3x – sin x]
\n= \\(\\frac{e^{x}}{2}\\) [6 cos 3x – 8 sin 3x – 2 cos x]
\n= ex<\/sup> [3 cos 3x – 4 sin 3x – cos x]<\/p>\n

\"Inter<\/p>\n

vi) Tan-1<\/sup>\\(\\frac{1+x}{1-x}\\)
\nSolution:
\ny = tan-1<\/sup>\\(\\frac{1+x}{1-x}\\)
\nPut x = tan \u03b8
\n\"Inter
\n\u2234 f(x) = \\(\\frac{\\pi}{4}\\) + tan-1<\/sup> (x)
\nDiff. w.r.to x
\nf'(x) = 0 + \\(\\frac{1}{1+x^{2}}\\)
\nf”(x) = (-1) (1 + x\u00b2)-2<\/sup> (2x)
\nf”(x) = \\(\\frac{-2 x}{\\left(1+x^{2}\\right)^{2}}\\)<\/p>\n

vii) tan-1<\/sup>\\(\\frac{3x-x^{3}}{1-3x^{2}}\\)
\nSolution:
\nf(x) = tan-1<\/sup>\\(\\frac{3x-x^{3}}{1-3x^{2}}\\) ; Put x = tan \u03b8
\nf(x) = tan-1<\/sup>\\(\\left(\\frac{3 \\tan \\theta-\\tan ^{3} \\theta}{1-3 \\tan ^{2} \\theta}\\right)\\)
\n= tan-1<\/sup>(tan 3\u03b8) = 3\u03b8
\n\u2234 f(x) = tan-1<\/sup> (x)
\nf'(x) = 3\\(\\frac{1}{1+x^{2}}\\) = \\(\\frac{3}{1+x^{2}}\\)
\nAgain Diff. w.r.to x
\nf”(x) = (3) (-1) (1 + x\u00b2)-2<\/sup> (2x)
\n\u21d2 f”(x) = \\(\\frac{-6 x}{\\left(1+x^{2}\\right)^{2}}\\)<\/p>\n

II. Prove the following.<\/span>
\nIf y = axn + 1<\/sup> + bx-n<\/sup>
\nthen x\u00b2y” = n(n + 1) y.
\nSolution:
\ny = axn + 1<\/sup> + bx-n<\/sup>
\ny1 = (n + 1). axn<\/sup> – n bx-n-1<\/sup>
\ny2 = n(n + 1). axn-1<\/sup> + n(n + 1) bx-n-2<\/sup>
\n\u2234 x\u00b2y2 = n(n + 1) axn+1<\/sup> + n(n + 1) bx-n<\/sup>
\n= n(n + 1) (axn+1+bx_n) = n(n + 1) y
\n\u2234 x\u00b2y” = n(n + 1) y<\/sub><\/p>\n

ii) If y = a cos x + (b + 2x) sin x, then y” + y = 4 cos x
\nSolution:
\n\\(\\frac{dy}{dx}\\) = y\u2019 = a(-sin x) + (b + 2x) \\(\\frac{d}{dx}\\) (sin x) + sin x \\(\\frac{d}{dx}\\) (b + 2x)
\n= – a sin x + (b + 2x) cos x + sin x.2
\n\\(\\frac{d^{2} y}{d x^{2}}\\) = y” = – a cos x + (b + 2x) (- sin x) + cos x (2) + 2 cos x
\nL.H.S. = y” + y = – a cos x + (b + 2x) (- sin x) + 2 cos x + 2 cos x + a cos x+ (b + 2x) sinx = 4 cosx<\/p>\n

iii) If y = 6 (x + 1) + (a + bx) e3x<\/sup> then y” – 6y’ + 9y = 54 x + 18
\nSolution:
\ny’ = 6 \\(\\frac{d}{dx}\\) (x + 1) + (a + bx) \\(\\frac{d}{dx}\\) (e3x<\/sup>) + e3x<\/sup> \\(\\frac{d}{dx}\\) (a + bx)
\n= 6(1) + (a + bx) 3e3x<\/sup> + e3x<\/sup>. b.
\ny” = 0 + 3 (a + bx) \\(\\frac{d}{dx}\\) e3x<\/sup> + 3e3x<\/sup> \\(\\frac{d}{dx}\\) (a + bx) + b \\(\\frac{d}{dx}\\) (e3x<\/sup>)
\n= 3(a + bx) 3e3x<\/sup> + 3e3x<\/sup>(b) + b. 3e3x<\/sup>
\n= 9(a + bx) e3x<\/sup> + 6b e3x<\/sup>
\nNow L.H.S. = y” – 6y’ + 9y = 9(a + bx) e3x<\/sup> + 6be3x<\/sup> – 6[6 + 3(a + bx)e3x<\/sup> + be3x<\/sup>] + 9 [6(x + 1) + (a + bx)e3x<\/sup>]
\n= 9(a + bx) e3x<\/sup> + 6b e3x<\/sup> – 36 – 18(a + bx)e3x<\/sup> – 6be3x<\/sup> + 54x + 54 + 9(a + bx) e3x<\/sup> = 54x + 18<\/p>\n

iv) If ay4<\/sup> = (x + b)5<\/sup> then 5y y” = (y’)\u00b2
\nSolution:
\nay4<\/sup> = (x + b)5<\/sup> ; y4<\/sup> = \\(\\frac{\\left(x+b\\right)^{2}}{a}\\)
\n\"Inter<\/p>\n

v) If y = a cos (sin x) + b sin (sin x) then y” + (tan x) y’ + y cos\u00b2 x = 0
\nSolution:
\ny = a cos (sin x) + b sin (sin x) …………. (1)
\nDifferentiating w.r.to x
\ny1<\/sub> = – a sin (sin x) cos x + b cos (sin x). cos x = [- a sin (sin x) + b cos (sin x)j cos x ………. (2)
\nDifferentiating again w.r.to x.
\ny2<\/sub> = – sin x [-a sin (sin x) + b cos (sin x)] + cos x [- a cos (sin x) cos x – b sin (sin x) cos x]
\n= – sin x. \\(\\frac{y_{1}}{\\cos x}\\) = -cos\u00b2 x.y
\nFrom (1) and (2),
\n\u21d2 y2<\/sub> + (tan x) y1<\/sub> + y cos\u00b2 x = 0<\/p>\n

\"Inter<\/p>\n

III.<\/span><\/p>\n

i) If y = 128 sin\u00b3 x cos4<\/sup> x, then find y”.
\nSolution:
\nf(x) = 128 sin\u00b3 x. cos4<\/sup> x
\nD.w.r. to x
\nf'(x)= 128 [sin\u00b3 x {4cos\u00b3 x (-sinx)} + cos4<\/sup> x {3sin\u00b2 x. cos x}]
\n= 128 [3sin\u00b2 x cos5<\/sup> x – 4sin4<\/sup> x cos\u00b3 x]
\nAgain D.w.r to x.
\nf”(x)= 128 {3 [sin\u00b2 x 5cos4<\/sup> x . (-sinx) + cos5<\/sup> x. 2sinx cosx] – 4 [sin4<\/sup> x. 3. cos2x (-sinx) + cos\u00b3 x. 4 sin\u00b3 x. cos x]}
\n= 128 [-15sin\u00b3 x . cos4<\/sup> x + 6sin x cos6<\/sup> x + 12 sin5<\/sup> x cos\u00b2 x- 16sin\u00b3 x cos4<\/sup> x]
\nf”(x)= 128 [6sinx cos6x + 12sin5<\/sup> x. cos\u00b2 x – 31sin\u00b3 x. cos4<\/sup> x]<\/p>\n

ii) If y = sin 2x sin 3x sin 4x, then find y”.
\nSolution:
\nf(x) = sin 2x sin 3x sin 4x
\n= \\(\\frac{1}{2}\\)sin 2x [2sin 4x. sin 3x]
\n= \\(\\frac{1}{2}\\)sin 2x [cos x – cos 7x]
\n= \\(\\frac{1}{2}\\) [sin 2x . cos x – cos 7x. sin 2x]
\n= \\(\\frac{1}{2}\\)\u00d7\\(\\frac{1}{2}\\) [2 sin2x. cosx – 2 cos7x. sin2x]
\n= \\(\\frac{1}{4}\\) [(sin3x + sinx) – (sin9x – sin5x)]
\n= \\(\\frac{1}{4}\\) [-sin9x + sin5x + sin3x + sinx]
\nD.w.r. to x
\nf'(x) = \\(\\frac{1}{4}\\) [-9 cos 9x + 5 cos 5x + 3 cos 3x + cos x]
\nD.w.r. to x
\nf”(x) = \\(\\frac{1}{4}\\) [81 sin 9x – 25sin 5x – 9sin 3x – sin x]<\/p>\n

iii) If ax\u00b2 + 2hxy + by\u00b2 = 1 then prove that
\n\\(\\frac{d^{2} y}{d x^{2}}=\\frac{h^{2}-a b}{(h x+b y)^{3}}\\)
\nSolution:
\nGiven ax\u00b2 + 2hxy + by\u00b2 = 1
\nDifferentiating w.r. to x
\na.2x + 2h(x. \\(\\frac{dy}{dx}\\) + y) + b . 2y\\(\\frac{dy}{dx}\\) = 0
\n2ax + 2hx. \\(\\frac{dy}{dx}\\) + 2hy + 2by. \\(\\frac{dy}{dx}\\) = 0
\n2(hx + by).\\(\\frac{dy}{dx}\\) = -2(ax + hy)
\n\\(\\frac{dy}{dx}\\) = \\(\\frac{-2(ax +hy)}{2(hx+by)}\\) = \\(\\frac{(ax+hy)}{(hx+by)}\\) ……… (1)
\nDifferentiating again w.r. to x, \\(\\frac{d^{2} y}{d x^{2}}\\)
\n\"Inter<\/p>\n

iv) If y = ae-bx<\/sup> cos(cx + d) then prove that y” + 2by’ + (b\u00b2 + c\u00b2) y = 0.
\nSolution:
\nGiven y = ae-bx<\/sup> cos(cx + d) ……….. (1)
\ny1<\/sub> = a[e-bx<\/sup> {- sin (cx + d)}. c. + cos (cx + d) e-bx<\/sup> (-b)}
\n= – a. e-bx<\/sup> [c sin (cx + d) + b cos (cx + d)]
\n= – ac.e-bx<\/sup> sin (cx + d) – by
\ny1<\/sub> + by = – ac. e-bx<\/sup> sin (cx + d) ………. (2)
\nDifferentiating once w.r.to x
\ny2<\/sub> + by1<\/sub> = – ac(- b) e-bx<\/sup> sin (cx + d) – ac e-bx<\/sup> cos (cx + d) (+ c)
\n= abce-bx<\/sup> sin (cx + d) -ac\u00b2 e-bx<\/sup> cos (cx + d)
\n= – b (y1<\/sub> + by) c\u00b2y [From (1) and (2)]
\n\u21d2 y2<\/sub> + by1<\/sub> + by1<\/sub> + b\u00b2y + c\u00b2y = 0.
\n\u21d2 y2<\/sub> + 2by1<\/sub> + (b\u00b2 + c\u00b2) y = 0 (or)
\ny” + 2 by’ + (b\u00b2 + c\u00b2) y = 0<\/p>\n

\"Inter<\/p>\n

v) If y = \\(e^{\\frac{-k}{2} x}\\) (a cos nx + b sin nx) prove that then y” + ky’+ (n\u00b2 + \\(\\frac{k^{2}}{4}\\))y = 0
\nSolution:
\nGiven y = \\(e^{\\frac{-k}{2} x}\\) (a cos nx + b sin nx) ………… (1)
\n\u2234 y1<\/sub> = \\(e^{\\frac{-k}{2} x}\\) (-n. a sin nx + n. b cos nx) + (a cos nx + b sin nx).\\(e^{\\frac{-k}{2} x}\\) (-\\(\\frac{k}{2}\\))
\n= –\\(\\frac{k}{2}\\) . y – n.e-kx\/2<\/sup> (a sin nx + b cos nx)
\n\u2234 y1<\/sub> + \\(\\frac{k}{2}\\) y = – n.e-kx\/2<\/sup> (a sin nx + b cos nx) …………. (2)
\nDifferentiating w.r. to x k
\ny2<\/sub> + \\(\\frac{k}{2}\\) y1<\/sub> = -n[{e-kx\/2<\/sup> (- na cos nx – nb sin nx)}] + {(a sin nx +b cos nx).e-kx\/2<\/sup> \\(\\frac{k}{2}\\)}
\n= – n\u00b2 e-kx\/2<\/sup>(a cos nx + b sin nx) – \\(\\frac{k}{2}\\){-n.e-kx\/2<\/sup> (a sin nx + b cos nx)}
\n\"Inter<\/p>\n","protected":false},"excerpt":{"rendered":"

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Differentiation Solutions Exercise 9(d) will help students to clear their doubts quickly. Intermediate 1st Year Maths 1B Differentiation Solutions Exercise 9(d) I. Question 1. If y = then find y” Solution: y = Differentiating w.r.to x Question 2. y = aenx …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[15],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35115"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=35115"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35115\/revisions"}],"predecessor-version":[{"id":35116,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35115\/revisions\/35116"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=35115"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=35115"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=35115"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}