NCERT Solutions for Class 12 Maths<\/a> Chapter 5 Continuity and Differentiability Ex 5.5 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-5-ex-5-5\/<\/p>\nNCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.5<\/h2>\n <\/p>\n
Ex 5.5 Class 12 NCERT Solutions Question 1.<\/strong> \ncos x. cos 2x. cos 3x \nSolution: \nLet y = cos x. cos 2x . cos 3x, \nTaking log on both sides, \nlog y = log (cos x. cos 2x. cos 3x) \nlog y = log cos x + log cos 2x + log cos 3x, \nDifferentiating w.r.t. x, we get \n <\/p>\nExercise 5.5 Class 12 Maths Solutions Question 2.<\/strong> \n\\(\\sqrt{\\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\\) \nSolution: \n <\/p>\n <\/p>\n
Ex 5.5 Class 12 Maths Ncert Question 3.<\/strong> \n(log x)cosx<\/sup> \nSolution: \nlet y = (log x)cosx<\/sup> \nTaking log on both sides, \nlog y = log (log x)cosx<\/sup> \nlog y = cos x log (log x), \nDifferentiating w.r.t. x, \n <\/p>\nExercise 5.5 Class 12 NCERT Solutions\u00a0 Question 4.<\/strong> \nx – 2sinx<\/sup> \nSolution: \nlet y = x – 2sinx<\/sup>, \n\u2234 y = u – v \nDifferentiating w.r.t. x, \n <\/p>\n <\/p>\n
Ch 5 Maths Class 12 Ex 5.5 NCERT Solutions Question 5.<\/strong> \n(x+3)2<\/sup>.(x + 4)3<\/sup>.(x + 5)4<\/sup> \nSolution: \nlet y = (x + 3)2<\/sup>.(x + 4)3<\/sup>.(x + 5)4<\/sup> \nTaking log on both side, \nlogy = log [(x + 3)2<\/sup> . (x + 4)3<\/sup> . (x + 5)4<\/sup>] \n= log (x + 3)2<\/sup> + log (x + 4)3<\/sup> + log (x + 5)4<\/sup> \nlog y = 2 log (x + 3) + 3 log (x + 4) + 4 log (x + 5) \nDifferentiating w.r.t. x, we get \n \nOn simplification, \n\\(\\frac { dy }{ dx }\\) = (x + 3)(x + 4)\u00b2(x + 5)\u00b3(9x\u00b2 + 70x + 133)<\/p>\nQuestion 6. \n\\({ \\left( x+\\frac { 1 }{ x } \\right) }^{ x }+{ x }^{ \\left( 1+\\frac { 1 }{ x } \\right) }\\) \nSolution: \nLet u = \\(\\left(x+\\frac{1}{x}\\right)^{x}\\) = \\(\\left(\\frac{x^{2}+1}{x}\\right)^{x}\\) \nv = \\(x^{\\left(1+\\frac{1}{x}\\right)}\\) \nLet y = u + v \nDifferentiating w.r.t. x, \n\u2234 \\(\\frac { dy }{ dx }\\) = \\(\\frac { du }{ dx }\\) + \\(\\frac { dv }{ dx }\\) … (1) \nu = \\(\\left(\\frac{x^{2}+1}{x}\\right)^{x}\\) \nTaking logaritham on both sides, \nlogu = x log \\(\\left(\\frac{x^{2}+1}{x}\\right)\\) \n\u2234 log u = x(log(x\u00b2+ 1) – logx) \nDifferentiating both sides w.r.t.x, \n \nv = \\(x^{\\left(1+\\frac{1}{x}\\right)}\\) \nTaking logaritham on both sides, \nlogv = (1 + \\(\\frac { 1 }{ x }\\)) log x \nDifferentiating both sides w.r.t. x, \n <\/p>\n
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Question 7. \n(log x)x<\/sup> + xlogx<\/sup> \nSolution: \nLet u = (log x)x<\/sup>, v = xlogx<\/sup> and y = u + v \nDifferentiating w.r.t. x, \n\u2234 \\(\\frac { dy }{ dx }\\) = \\(\\frac { du }{ dx }\\) + \\(\\frac { dv }{ dx }\\) … (1) \nu = (log x)x<\/sup> \nTaking logaritham on both sides, \nlogu = x log(log x) \nDifferentiating both sides w.r.t. x, \n \nv = xlog x<\/sup> \nTaking logarithm on both sides, \nlogv = logx. logx = (logx)\u00b2 \nDifferentiating both sides w.r.t. x, \n <\/p>\nQuestion 8. \n(sin x)x<\/sup>+sin-1<\/sup> \\(\\sqrt{x}\\) \nSolution: \nLet y = (sin x)x\u00a0<\/sup>+ sin-1\u00a0<\/sup>\\(\\sqrt{x}\\) \nLet y = u + v \nDifferentiating w.r.t. x, \n\u2234 \\(\\frac { dy }{ dx }\\) = \\(\\frac { du }{ dx }\\) + \\(\\frac { dv }{ dx }\\) … (1) \nu = (sin x)x<\/sup> \n\\(\\frac { du }{ dx }\\) = (sin x)x<\/sup>[x cotx + log sinx] … (2) \nLet y = (sinx)x<\/sup> \nTaking logarithm on bath sides, \nlogy = x.log sin x \nDifferentiating both sides w.r.t. x, \n <\/p>\n <\/p>\n
Question 9. \nxsinx<\/sup> + (sin x)cosx<\/sup> \nSolution: \nLet y = xsinx<\/sup> + (sin x)cosx<\/sup> \nDifferentiating both sides w.r.t. x, \n \n(i) Let y = (sin x)x<\/sup> \nTaking logarithm on both sides, \nlogy = sinx. logx \nDifferentiating both sides w.r.t. x, \n <\/p>\n(ii) Let y = (sinx)cosx\u00a0<\/sup> \nTaking logarithm on both sides, \nlogy = cosx. log sinx \nDifferentiating both sides w.r.t. x, \n <\/p>\nQuestion 10. \n\\({ x }^{ x\\quad cosx }+\\frac { { x }^{ 2 }+1 }{ { x }^{ 2 }-1 } \\) \nSolution: \nLet u = xx cosx<\/sup> and v = \\(\\frac{x^{2}+1}{x^{2}-1}\\) \nLet y = u + v \nDifferentiating w.r.t. x, \n\u2234 \\(\\frac { dy }{ dx }\\) = \\(\\frac { du }{ dx }\\) + \\(\\frac { dv }{ dx }\\) … (1) \nu = xx cosx<\/sup> \nTaking logarithm on both sides, \nlogu = x cosx . logx \nDifferentiating both sides w.r.t. x, \n \nDifferentiating both sides w.r.t. x, \n <\/p>\n <\/p>\n
Question 11. \n\\((x \\cos x)^{x}+(x \\sin x)^{\\frac{1}{x}}\\) \nSolution: \nLet u = (x cosx)x<\/sup> and v = \\((x \\sin x)^{\\frac{1}{x}}\\) \nu = (x cosx)x<\/sup> \nTaking logarithm on both sides, \nlog u = x log (x cosx) \nlog u = x[log x + log cosx] \nDifferentiating w.r.t. x, \n <\/p>\nQuestion 12. \nxy<\/sup> + yx<\/sup> = 1 \nSolution: \nxy<\/sup> + yx<\/sup> = 1 \nlet u = xy<\/sup> and v = yx<\/sup> \n\u2234 u + v = 1, \n\\(\\frac { du }{ dx } +\\frac { dv }{ dx }=0\\) \nNow u = xy<\/sup> \nTaking logarithm on both sides, \nlog u = y log x \nDifferentiating w.r.t. x \n <\/p>\n <\/p>\n
Question 13. \nyx<\/sup> = xy<\/sup> \nSolution: \nxy<\/sup> = yx<\/sup> \nTaking logarithm on both sides, \ny log x = x log y \nDifferentiating w.r.t. x \n <\/p>\nQuestion 14. \n(cos x)y<\/sup> = (cos y)x<\/sup> \nSolution: \n(cos x)y<\/sup> = (cos y)x<\/sup> \nTaking logarithm on both sides, \ny log cosx = x log cosy \nDifferentiating both sides w.r.t. x, \n <\/p>\nQuestion 15. \nxy = e(x-y)<\/sup> \nSolution: \nlog(xy) = log e(x-y)<\/sup> \n\u21d2 log(xy) = x – y \n\u21d2 logx + logy = x – y \n\\( \u21d2\\frac { 1 }{ x } +\\frac { 1 }{ y } \\frac { dy }{ dx } =1-\\frac { dy }{ dx } \u21d2\\frac { dy }{ dx } =\\frac { y(x-1) }{ x(y+1) } \\)<\/p>\nQuestion 16. \nFind the derivative of the function given by f (x) = (1 + x) (1 + x2<\/sup>) (1 + x4<\/sup>) (1 + x8<\/sup>) and hence find f'(1). \nSolution: \nLet f(x) = y = (1 + x)(1 + x2<\/sup>)(1 + x4<\/sup>)(1 + x8<\/sup>) \nTaking log both sides, we get \nlogy = log [(1 + x)(1 + x2<\/sup>)(1 + x4<\/sup>)(1 + x8<\/sup>)] \nlogy = log(1 + x) + log (1 + x2<\/sup>) + log(1 + x4<\/sup>) + log(1 + x8<\/sup>) \nDifferentiating w.r.t. x, \n <\/p>\n <\/p>\n
Question 17. \nDifferentiate (x2<\/sup> – 5x + 8) (x3<\/sup> + 7x + 9) in three ways mentioned below: \n(i) by using product rule \n(ii) by expanding the product to obtain a single polynomial. \n(iii) by logarithmic differentiation. \nDo they all give the same answer? \nSolution: \n(i) By using product rule \nf’ = (x2<\/sup> – 5x + 8) (3x2<\/sup> + 7) + (x3<\/sup> + 7x + 9) (2x – 5) \nf = 5x4<\/sup> – 20x3<\/sup> + 45x2<\/sup> – 52x + 11.<\/p>\n(ii) y = x5<\/sup> – 5x4<\/sup> + 15x\u00b3 – 26x\u00b2 + 11x + 72 \n(by expanding the product) \nDifferentiating w.r.t. x, dy \n\\(\\frac { dy }{ dx }\\) = 5x4<\/sup> – 20x\u00b3 + 45x\u00b2 – 52x + 11<\/p>\n(iii) y = (x\u00b2 – 5x + 8)(x\u00b3 + 7x + 9) \nTaking logarithm on both sides, \nlog y = log(x\u00b2 – 5x 4- 8)(x\u00b3 + 7x + 9) \nlog y = log(x\u00b2 – 5x + 8) + log(x\u00b3 + 7x + 9) \nDifferentiating, w.r.t x, \n \n= (2x – 5) (x\u00b3 + 7x + 9) + (3x\u00b2 + 7) (x\u00b2 – 5x + 8) \n= 5x4<\/sup> – 20x\u00b3 + 45x\u00b2 – 52x + 11 \nYes, the answers are the same.<\/p>\n <\/p>\n
Question 18. \nIf u, v and w are functions of w then show that \n\\(\\frac { d }{ dx } (u.v.w)=\\frac { du }{ dx } v.w+u.\\frac { dv }{ dx } .w+u.v\\frac { dw }{ dx } \\) \nin two ways-first by repeated application of product rule, second by logarithmic differentiation. \nSolution: \nLet y = u.v.w \n(a) Product rule \n <\/p>\n
(b) Logarithmic differentiation \ny = uvw \nTaking logarithm on both sides, we get \nlog y = log uvw \ni.e., log y = log u + log v + log w \nDifferentiating both sides w.r.t. x, we get \n <\/p>\n","protected":false},"excerpt":{"rendered":"
These NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-5-ex-5-5\/ NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.5 Ex 5.5 Class 12 NCERT Solutions Question 1. cos x. cos 2x. cos 3x Solution: Let …<\/p>\n
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[3],"tags":[],"yoast_head":"\nNCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5 - MCQ Questions<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n