{"id":6299,"date":"2024-01-11T02:50:08","date_gmt":"2024-01-10T21:20:08","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=6299"},"modified":"2024-01-12T17:50:04","modified_gmt":"2024-01-12T12:20:04","slug":"ts-inter-1st-year-maths-1b-solutions-chapter-9-ex-9a","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-inter-1st-year-maths-1b-solutions-chapter-9-ex-9a\/","title":{"rendered":"TS Inter 1st Year Maths 1B Solutions Chapter 9 Differentiation Ex 9(a)"},"content":{"rendered":"

Students must practice these TS Inter 1st Year Maths 1B Study Material<\/a> Chapter 9 Differentiation Ex 9(a) to find a better approach to solving the problems.<\/p>\n

TS Inter 1st Year Maths 1B Differentiation 9(a)<\/h2>\n

I.
\nQuestion 1.
\nFind the derivatives of the following functions f(x). (V.S.A.Q.)
\n(i) \u221ax + 2x3\/4<\/sup> + 3x5\/6<\/sup> (x > 0)
\nAnswer:
\n\"TS<\/p>\n

\"TS<\/p>\n

(ii) \\(\\sqrt{2 x-3}+\\sqrt{7-3 x}\\)
\nAnswer:
\n\"TS<\/p>\n

(iii) (x2<\/sup> – 3) (4x3<\/sup> + 1)
\nAnswer:
\ny = (x2<\/sup> – 3) (4x3<\/sup> + 1)
\n\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = (x2<\/sup> – 3) \\(\\frac{\\mathrm{d}}{\\mathrm{dx}}\\) (4x3<\/sup> + 1) + (4x3<\/sup> + 1) \\(\\frac{\\mathrm{d}}{\\mathrm{dx}}\\) (x2<\/sup> – 3)
\n= (x2<\/sup> – 3) (12x2<\/sup>) + (4x3<\/sup> + 1) (2x)
\n= 12x4<\/sup> – 36x2<\/sup> + 8x4<\/sup> + 2x = 20x4<\/sup> – 36x2<\/sup> + 2x<\/p>\n

(iv) (\u221ax – 3x) (x + \\(\\frac{1}{x}\\))
\nAnswer:
\n\"TS<\/p>\n

(v) (\u221ax + 1) (x2<\/sup> – 4x + 2) (x > 0)
\nAnswer:
\ny = (\u221ax + 1) (x2<\/sup> – 4x + 2)
\n\\(\\frac{d y}{d x}\\) = (\u221ax + 1) \\(\\frac{d}{d x}\\) (x2<\/sup> – 4x + 2) + (x2<\/sup> – 4x + 2) \\(\\frac{d}{d x}\\) (\u221ax + 1)
\n= (\u221ax + 1) (2x – 4) + (x2<\/sup> – 4x + 2) \\(\\left(\\frac{1}{2 \\sqrt{x}}\\right)\\)
\n= (\u221ax + 1) (2x – 4) + \\(\\frac{x^2-4 x+2}{2 \\sqrt{x}}\\)<\/p>\n

(vi) (ax + b)n<\/sup> (cx + d)m<\/sup>
\nAnswer:
\nLet y = (ax + b)n<\/sup> (cx + d)m<\/sup>
\n\\(\\frac{d y}{d x}\\) = (ax + b)n<\/sup> . m(cx + d)m – 1<\/sup>(c) + (cx + d)m<\/sup> n (ax + b)n – 1<\/sup> (a)
\n= (ax + b)n<\/sup> (cm (cx + d)m – 1<\/sup>] + (cx + d)m<\/sup> [an(ax + b)n – 1<\/sup>]
\n= (ax + b)n – 1<\/sup> (cx + d)m – 1<\/sup> [cm (ax + b) + an(cx + d)]
\n= (ax + b)n<\/sup> (cx + d)m<\/sup> \\(\\left[\\frac{a n}{a x+b}+\\frac{c m}{c x+d}\\right]\\)<\/p>\n

(vii) 5 sin x + ex<\/sup> log x
\nAnswer:
\ny = 5 sin x + ex<\/sup> log x
\n\"TS<\/p>\n

\"TS<\/p>\n

(viii) 5x<\/sup> + log x + x3<\/sup> ex<\/sup>
\nAnswer:
\ny = 5x<\/sup> + log x + x3<\/sup> ex<\/sup>
\n\u2234 \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = 5x<\/sup> log 5 + \\(\\frac{1}{x}\\) + [x3<\/sup>ex<\/sup> + 3x2<\/sup>ex<\/sup>]<\/p>\n

(ix) ex<\/sup> + sin x cos x
\nAnswer:
\ny = ex<\/sup> + sin x cos x
\n\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = ex<\/sup> + sin x (- sin x) + cos x (cos x) dx
\n= ex<\/sup> + cos2<\/sup> x – sin2<\/sup> x = ex<\/sup> + cos 2x<\/p>\n

(x) \\(\\frac{p x^2+q x+r}{a x+b}\\) (|a| + |b| \u2260 0)
\nAnswer:
\n\"TS<\/p>\n

(xi) log7<\/sub>(log x) (x > 0)
\nAnswer:
\ny = log7<\/sub>(log x)
\n\"TS<\/p>\n

(xii) \\(\\frac{1}{a x^2+b x+c}\\) (|a| + |b| + |c| \u2260 0)
\nAnswer:
\n\"TS<\/p>\n

(xiii) e2x<\/sup> log (3x + 4) (x > – 4\/3)
\nAnswer:
\ny = e2x<\/sup> log (3x + 4) (x > – 4\/3)
\n\"TS<\/p>\n

\"TS<\/p>\n

(xiv) (4 + x2<\/sup>) e2x<\/sup>
\nAnswer:
\ny = (4 + x2<\/sup>) e2x<\/sup>
\n\u2234 \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = (4 + x2<\/sup>) \\(\\frac{\\mathrm{d}}{\\mathrm{dx}}\\) (e2x<\/sup>) + e2x<\/sup> . \\(\\frac{\\mathrm{d}}{\\mathrm{dx}}\\) (4 + x2<\/sup>)
\n= (4 + x2<\/sup>) (2e2x<\/sup>) + e2x<\/sup> (2x)
\n= 2e2x<\/sup> (x2<\/sup> + x + 4)<\/p>\n

(xv) \\(\\frac{a x+b}{c x+d}\\) (|c| + |d| \u2260 0) (May 2022)
\nAnswer:
\ny = \\(\\frac{a x+b}{c x+d}\\)
\n\"TS<\/p>\n

(xvi) ax<\/sup>.ex2<\/sup><\/sup> (Board New Model Paper)
\nAnswer:
\ny = ax<\/sup>.ex2<\/sup><\/sup>
\nThen
\n\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = ax<\/sup> . \\(\\frac{\\mathrm{d}}{\\mathrm{dx}}\\) (ex2<\/sup><\/sup>) + \\(\\frac{\\mathrm{d}}{\\mathrm{dx}}\\) (ax<\/sup>)
\n= ax<\/sup> (2x) ex2<\/sup><\/sup> + ex2<\/sup><\/sup> . ax<\/sup> loge<\/sub>a
\n= ax<\/sup> ex2<\/sup><\/sup> (2x + loge<\/sub>a<\/sup>)
\n= y (2x + loge<\/sub>a<\/sup>)<\/p>\n

Question 2.
\nIf f(x) = (1 + x + x2<\/sup> + …… + x100<\/sup>) then find f ‘(1). (V.S.A.Q.)
\nAnswer:
\nf(x) = (1 + x + x2<\/sup> + ……. + x100<\/sup>)
\nThen f(x) = 1 + 2x+ 3x2<\/sup> + …………….. + 100x99<\/sup>
\nand f\u2019(1) = 1 + 2 + 3 + ………………. + 100
\n= \\(\\frac{100(100+1)}{2}\\) = 101 \u00d7 50 = 5050
\n(\u2235 \u03a3n = \\(\\frac{\\mathrm{n}(\\mathrm{n}+1)}{2}\\))<\/p>\n

Question 3.
\nIf f(x) = 2x + 3x – 5 then prove that f ‘(0) + 3f ‘( -1) = 0. (V.S.A.Q.)
\nAnswer:
\nGiven f(x) = 2x2<\/sup> + 3x – 5
\nThen f'(x) = 4x + 3
\n\u2234 f(0) + 3f (- 1) = 3 + 3 [4 (- 1) + 3] = 3 – 3 = 0<\/p>\n

\"TS<\/p>\n

II.
\nQuestion 1.
\nFind the derivatives of the following functions from the first principle. (S.A.Q.)
\n(i) x3<\/sup>
\nAnswer:
\nLet f(x) = x3<\/sup> then
\n\"TS<\/p>\n

(ii) x4<\/sup> + 4
\nAnswer:
\nlet f(x) = x4<\/sup> + 4
\n\"TS<\/p>\n

(iii) ax2<\/sup> + bx + c
\nAnswer:
\nLet f(x) = ax2<\/sup> + bx + c
\n\"TS<\/p>\n

\"TS<\/p>\n

(iv) \u221ax + 1
\nAnswer:
\nLet f(x) = \u221ax + 1
\n\"TS<\/p>\n

(v) sin 2x (Board New Model Paper)
\nAnswer:
\nLet f(x) = sin 2x
\n\"TS<\/p>\n

(vi) cos ax (May 2014, March 2013, 2011)
\nAnswer:
\nLet f(x) = cos ax
\n\"TS<\/p>\n

\"TS<\/p>\n

(vii) tan 2x (March 2014, May 2011)
\nAnswer:
\nLet f(x) = tan 2x
\n\"TS<\/p>\n

(viii) cot x
\nAnswer:
\nlet f(x) = cot x
\nthen
\n\"TS<\/p>\n

(ix) sec 3x
\nAnswer:
\nLet f(x) = sec 3x
\nthen
\n\"TS<\/p>\n

\"TS<\/p>\n

(x) x sin x
\nAnswer:
\nLet f(x) = x sin x
\nThen f(x + h) – f(x) = (x + h) sin (x + h) – x sin x
\n= x [sin(x + h) – sin x] + h sin (x + h)
\n\"TS
\n= x cos x \u00d7 1 + sin x
\n= x cos x + sin x<\/p>\n

(xi) cos2<\/sup>x
\nAnswer:
\nLet f(x) = cos2<\/sup>x
\nThen f (x + h) – f(x) = cos2<\/sup> (x + h) – cos2<\/sup>x
\n= [1 – sin2<\/sup>(x + h)] – (1 – sin2<\/sup>x)
\n= sin2<\/sup>x – sin2<\/sup>(x + h)
\n= sin (x + x + h) sin (x – x – h)
\n= sin (2x + h) sin (- h)
\n\"TS
\n= – sin 2x \u00d7 1 = – sin 2x<\/p>\n

Question 2.
\nFind the derivatives of the following functions. (S.A.Q.)
\n(i) \\(\\frac{1-x \\sqrt{x}}{1+x \\sqrt{x}}\\) (x > 0)
\nAnswer:
\n\"TS<\/p>\n

\"TS<\/p>\n

(ii) xn<\/sup> nx<\/sup> log (nx), (x > 0, n \u2208 N)
\nAnswer:
\nLet y = xn<\/sup> nx<\/sup> log (nx)
\nThen
\n\"TS<\/p>\n

(iii) ax2n<\/sup> log x + bxn<\/sup> e-x<\/sup>
\nAnswer:
\nLet y = ax2n<\/sup> log x + bxn<\/sup> e-x<\/sup>
\n\"TS
\n= a [x2n – 1<\/sup> + 2nx2n – 1<\/sup> log x] + b [- xn<\/sup> e-x<\/sup> + ne-x<\/sup> xn – 1<\/sup>]
\n= ax2n – 1<\/sup> + 2anx2n – 1<\/sup> log x – bxn<\/sup>e-x<\/sup> + bne-x<\/sup> xn – 1<\/sup><\/p>\n

(iv) \\(\\left(\\frac{1}{x}-x\\right)^3\\) ex<\/sup>
\nAnswer:
\n\"TS<\/p>\n

Question 3.
\nShow that the function f(x) = | x | + | x – 1| x \u2208 R is differentiable for all real numbers except for 0 and 1. (S.A.Q.)
\nAnswer:
\nf(x) = |x| + |x – 1| \u2200 x \u2208 R
\nthen f(x) = x + x – 1 = 2x – 1, x \u2265 1
\n= x – (x – 1) = x – x + 1 = 1, 0 < x < 1
\n= – x – (x – 1)
\n= – x – x + 1 = 1 – 2x; x \u2264 0
\n\u2234 f(x) = 2x – 1, x \u2265 1
\n= 1, 0 < x < 1 = 1 – 2x; x \u2264 0 If x > 1, then f(x) = 2x – 1 which is a polynomial differentiable for all x > 1.
\nIf 0 < x \u2264 1, then f(x) = 1 = Constant which is differentiable.<\/p>\n

Case (i): Differentiability at x = 0:
\n\"TS
\n\u2235 Lf'(0) \u2260 Rf'(0), the function f is not differentiable at x = 0.<\/p>\n

Case (i): Differentiability at x = 1:
\n\"TS
\n\u2235 Lf'(1) \u2260 Rf'(1) we say that f(x) is not differentiable at x = 1.
\n\u2234 f(x) is differentiable for all real x except zero and one.<\/p>\n

\"TS<\/p>\n

Question 4.
\nVerify whether the following function is differentiable at 1 and 3. (S.A.Q.)
\n\"TS
\nAnswer:
\nCase (i):
\n\"TS<\/p>\n

Case (ii):
\n\"TS
\nSince Lf'(3) \u2260 Rf'(3), f(x) is not differentiable at x = 3.<\/p>\n

Question 5.
\nIs the following function f differentiable at 2 ? Justify. (S.A.Q.)
\n\"TS
\nAnswer:
\n\"TS
\nSince Lf'(2) \u2260 Rf'(2), we say f(x) is not differentiable at x = 2.<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice these TS Inter 1st Year Maths 1B Study Material Chapter 9 Differentiation Ex 9(a) to find a better approach to solving the problems. TS Inter 1st Year Maths 1B Differentiation 9(a) I. Question 1. Find the derivatives of the following functions f(x). (V.S.A.Q.) (i) \u221ax + 2×3\/4 + 3×5\/6 (x > 0) … 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\/6299"}],"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=6299"}],"version-history":[{"count":3,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/6299\/revisions"}],"predecessor-version":[{"id":6456,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/6299\/revisions\/6456"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=6299"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=6299"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=6299"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}