math prblm derivtv

math prblm derivtv - Derivative a<x—1(2x3—1 2x2 ax 1...

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Unformatted text preview: Derivative a <x—1)(2x3—1} _ 2x2+ax+1 E X3 7 1 {x2 +x+ 1}a Fwssibie derivmiun: d (1—1.1(212—1] 3 13—1 Usethequotiemrlfle, d—[EJ=M, whereu=(x—1}[2xz —1) and V V F“ v=x3 —'l: {x3 —1}(:{{x—1;{2x3—1}]}—(x—1){2x3—1}(:(x3 —1}} (x 71} Usathepruductrule, —(u\)—\%+u :—." WhETELl—X— landv—sz —1: r \ =(x3:1}2[(" ‘9“sz 1}[%(r—1)]+(x—1)[%{2x2—1}]]_ (x—1){2x3—1][%{x3 _1]D Differentiatethe Sumten'n byterm: ={x3i1}2[(x371}[(x 1%; {2x3 1}]+ +{2x271}[%fx71]]]7 (x—1){2x3—1][%{x3]+%(_1}]] Differentiatethe Sumten'n byterm: [{x3—1}[(x— 1%; {2x3—1}]+ {2x3—1}[%{—1]+£(x]]]— (x— 1){2r3 — 1] [£98] + £91)” Theda-imtiveof—‘l iszem: Thedenvutweofil isaero: = {x51} [ix —1}[(x—1i[%{2x3—1}]+{2x3—1}[%m+o]]_ (x— 1H2)? — 1] [iifln Thedenvutweofxg 15 3x2: {x3 —1}((x—1}{:{2x3—1}]+{2x3—1]{:(x)}}—(x—1}(2x3—1]{3x3} (x 71} Theda-ivativeofxis 1: (x3 —1}((x— 1H} {23.3— 1}]+1{2x3 —1}]— 3(x— 1ix3{2x3— 1} ix -1} Different ate the sum harm by term and factur out constants {x371}((x71}{2{i{x1}]+i(71]}+2x271}73(x71)x1[2x271] bra—1r Theda‘ivativeof—liszero: {x3—1}((x—1}{2{i{x3}]+n}+2x3 —1}—3(x—1}x3{2x3—1} (xi—1}: Thedenvatweofx2152x: {x3—1}[2x3+2(x—1){2x)—1}—3:x—1]x3[2x3—1} {x3 71? EtIVE of ((x—l:(2x"2 —1]JJ(x"3 —1] ...
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