y = tan−1[log(e/x3)log(ex3)]+tan−1[3+2logx)1−6logx)] show that d2ydx2=0
simplify,f(x)=tan−1{logee−logex2logee+logex2}+tan−1{3+2logx1−6logx}f(x)=tan−1{1−2lagx1+2logx}+tan−1{3+2logx1−6logx}∣∣∣suppose{1−2lagx1+2logx}=a,{3+2logx1−6logx}=b⇒tan−1a+tan−1b=tan−1(a+b1−ab)wefindthevalueofa&b.a=1−2logx1+2logxandb=3+2logx1−6logxa+b=1−2logx1+2logx+3+2logx1−6logxa+b=(1−2logx)(1−6logx)+(3+2logx)(1+2logx)(1+2logx)(1−6logx)a+b=1−6logx−2logx+12(logx)2+3+6logx+2logx+4(logx)2(1+2logx)(1−6logx)a+b=4+16(logx)2(1+2logx)(1−6logx),wegetvalueof:a+b.Again,findthevalue(1−ab)(1−ab)=1−(1−2logx1+2logx)(3+2logx1−6logx)=(1+2logx)(1−6logx)−(1−2logx)(3−2logx)(1+2logx)(1−6logx)=1−6logx+2logx−12(logx)2−(3+2logx−6logx−4(logx)2(1+2logx)(1−6logx)=1−6logx+2logx−12(logx)2−3−2logx+6logx+4(logx)2(1+2logx)(1−6logx)∴1−ab=−2−8(logx)2(1+2logx)(1−6logx)|wegetthevalueof1−ab.wehave,tan−1(a+b1−ab)⇒tan−1⎛⎜
⎜
⎜
⎜
⎜⎝4+16(logx)2(1+2logx)(1−6logx)−2−8(logx)2(1+2logx)(1−6logx)⎞⎟
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⎟
⎟
⎟⎠⇒tan−1(4(1+4(logx)2−2(1+4(logx)2)⇒tan−1(−2)Finallyweget:f(x)=tan−1(−2)Now,dnydxn=Differentietof.......f′(x)=0f′′(x)=0...fn(x)=0So,thatthevalueofdnydxn=0
simplify,f(x)=tan−1{logee−logex2logee+logex2}+tan−1{3+2logx1−6logx}f(x)=tan−1{1−2lagx1+2logx}+tan−1{3+2logx1−6logx}∣∣∣suppose{1−2lagx1+2logx}=a,{3+2logx1−6logx}=b⇒tan−1a+tan−1b=tan−1(a+b1−ab)wefindthevalueofa&b.a=1−2logx1+2logxandb=3+2logx1−6logxa+b=1−2logx1+2logx+3+2logx1−6logxa+b=(1−2logx)(1−6logx)+(3+2logx)(1+2logx)(1+2logx)(1−6logx)a+b=1−6logx−2logx+12(logx)2+3+6logx+2logx+4(logx)2(1+2logx)(1−6logx)a+b=4+16(logx)2(1+2logx)(1−6logx),wegetvalueof:a+b.Again,findthevalue(1−ab)(1−ab)=1−(1−2logx1+2logx)(3+2logx1−6logx)=(1+2logx)(1−6logx)−(1−2logx)(3−2logx)(1+2logx)(1−6logx)=1−6logx+2logx−12(logx)2−(3+2logx−6logx−4(logx)2(1+2logx)(1−6logx)=1−6logx+2logx−12(logx)2−3−2logx+6logx+4(logx)2(1+2logx)(1−6logx)∴1−ab=−2−8(logx)2(1+2logx)(1−6logx)|wegetthevalueof1−ab.wehave,tan−1(a+b1−ab)⇒tan−1⎛⎜ ⎜ ⎜ ⎜ ⎜⎝4+16(logx)2(1+2logx)(1−6logx)−2−8(logx)2(1+2logx)(1−6logx)⎞⎟ ⎟ ⎟ ⎟ ⎟⎠⇒tan−1(4(1+4(logx)2−2(1+4(logx)2)⇒tan−1(−2)Finallyweget:f(x)=tan−1(−2)Now,dnydxn=Differentietof.......f′(x)=0f′′(x)=0...fn(x)=0So,thatthevalueofdnydxn=0
