A Text Book of Engineering Mathematics. Volume I by Rajesh Pandey

By Rajesh Pandey

Quantity i of this sequence serves as a textbook for semester i of thesubject engineering arithmetic. compatible figures and diagrams havebeen used to make sure a simple knowing of the strategies concerned. to stress software of the subjects mentioned, compatible examplesare included in the course of the ebook. Solved examples, within the bookinclude recommendations of questions from past u. P. T. U. Examinations. earlier years query papers were incorporated as to exposestudents to the trend and kind of questions they might face in anexamination. This ebook is especially valuable for measure, honours andpostgraduate scholars of all indian universities and for ias, pcsand different aggressive examinations. in regards to the writer dr. Rajesh pandey he has greater than fourteen years experiencein instructing scholars of undergraduate, postgraduate and engineeringlevel. He acquired his b. Sc and m. Sc measure in 1991 and 1993respectively from gorakhpur collage and phd in 12 months 2003 andparticipated in quite a few seminars & meetings of nationwide andinternational point. For graduate & postgraduate, the authorhas additionally written books on boost calculus, vectors, numericalanalysis, summary algebra, mechanics, fluid mechanics and so forth. shortly, he's operating as an assistant professor/reader inmathematics, sherwood collage of engineering, learn andtechnology, lucknow. desk of contents uncomplicated effects and ideas successive differentiation and leibnitz's theorem partial differentiation curve tracing enlargement of functionality jacobian approximation of blunders extrema of services of numerous variables lagranges approach to undetermind multipliers matrices a number of integers beta and gamma capabilities vector differential calculus vector quintessential calculus exam papers of uptu from 2001-2009

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In (v), we get (yn)o = - (n - 2)2 (y n- 2)0; (yn- 2)0 = - (n - 4)2 (yn -4)0; (Yn-4)0 = - (n - 6)2(Yn_6)0; where we have (on multiplying side wise) (yn)O = {-(n - 2)2} {- (n - 4)2} { -(n - 6)2} {(yn -6)0} :. If n is odd, (yn)O = {- (n - 2)2}{ - (n - 4)2} { - (n - 6)2} .................. { -32} { -12}(Yt)o Now from (i) putting x =0, we get (y)o = [log (0+1)]2 = (log 1)2 =0 :. From (ii) putting x =0, we get (Yl)~ = 4(y)0 =0 or (yt)o =0 Hence when n is odd (Yn)O =0 Answer And if n is even, as before (yn)o = { -l(n - 2)2}{ -(n - 4)2} { -(n - 6)2} ................

V. ;------'~ (l+logz) (ii) Similarly from (i) we have az (l+logy) ay =-(l+logz) :. [-(l + logzt..!.. {_(1 + log x)), using (ii) axay z(l+logz) l+logz a2z (1 +logx)2 At x = y = z, we have - - = 3 axOy x(1+1ogx) Substituting x for y and z . a2 z 1 1 e - - =-~----:.. log e = 1 =- x(loge'+logx) 1 =---- x log (ex) = - {x log (ex)}-l Hence Proved. 2006) . Weave h ' -1 So Iuhon u- sm (x)y + tan (r)x' ............ t. x and y, we get. au Ox = rn + 1 oy1 (-y) 1 1+(~)' 0 -;1 24 Partial Differentiation = x x +-::--7" y ~y2 _ X2 X2 + y2 au or x ay x xy =- ~y2 _ x2 + x2 + y2 .................

A2 z 1 1 e - - =-~----:.. log e = 1 =- x(loge'+logx) 1 =---- x log (ex) = - {x log (ex)}-l Hence Proved. 2006) . Weave h ' -1 So Iuhon u- sm (x)y + tan (r)x' ............ t. x and y, we get. au Ox = rn + 1 oy1 (-y) 1 1+(~)' 0 -;1 24 Partial Differentiation = x x +-::--7" y ~y2 _ X2 X2 + y2 au or x ay x xy =- ~y2 _ x2 + x2 + y2 ................. (... ) 111 on adding (ii) and (iii), we have au au x-+y-=O Answer. V. V. 2004) Proof: Since f(x,y) is a homogeneous function of degree n, it can be expressed in the form f(x,y) = xn F(y Ix) ...

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