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Question-173834

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Question Number 173834 by mnjuly1970 last updated on 19/Jul/22
Commented by infinityaction last updated on 20/Jul/22
f(x)g(x)h(x) = (tan^7 x+cot^7 x)(tan^8 x−cot^8 x) × (tan^9 x+cot^9 x) F(x) = (tan^(15) x−cotx+tanx−cot^(15) x)(tan^9 +cot^9 x) F(x)= tan^(24) x−tan^8 x+tan^(10) x−cot^6 x + tan^6 x−cot^(10) x+cot^8 x−cot^(24) x F ′(x) = 24tan^(23) xsec^2 x−8tan^7 xsec^2 x + 10tan^9 xsec^2 x+6cot^5 xcosec^2 x + 6tan^5 xsec^2 x+10cot^9 xcosec^2 x− 8cot^7 xcosec^2 x +24cot^(23) xcosec^2 x F ′(x) = 48−16+20+12+12+20−16+48 F^ ′(x) = 128
f(x)g(x)h(x)=(tan7x+cot7x)(tan8x−cot8x)×(tan9x+cot9x)F(x)=(tan15x−cotx+tanx−cot15x)(tan9+cot9x)F(x)=tan24x−tan8x+tan10x−cot6x+tan6x−cot10x+cot8x−cot24xF′(x)=24tan23xsec2x−8tan7xsec2x+10tan9xsec2x+6cot5xcosec2x+6tan5xsec2x+10cot9xcosec2x−8cot7xcosec2x+24cot23xcosec2xF′(x)=48−16+20+12+12+20−16+48Prime causes double exponent: use braces to clarify
Commented by Tawa11 last updated on 19/Jul/22
Great sirs
Greatsirs
Answered by mahdipoor last updated on 19/Jul/22
P (x)=P_(a,±) (x)=tan^a (x)±cot^a (x) (dP/dx)=a(tan^(a−1) (x).(1/(cos^2 x))∓cot^(a−1) (x).(1/(sin^2 x))) ⇒ { ((P(θ)=1±1)),((P^( ′) (θ)=2a(1∓1))) :} get θ=π/4 f ≡ P_(7,+) ⇒f(θ)=2 D_x f(θ)=0 g ≡ P_(8,−) ⇒g(θ)=0 D_x g(θ)=32 h ≡ P_(7,+) ⇒h(θ)=2 D_x h(θ)=0 (dF/dx)(θ)=(d/dx)(fgh)(θ)=(f^( ′) gh+fg^( ′) h+fgh^( ′) )(θ)= 0+2×32×2+0=128
P(x)=Pa,±(x)=tana(x)±cota(x)dPdx=a(tana−1(x).1cos2x∓cota−1(x).1sin2x)⇒{P(θ)=1±1P′(θ)=2a(1∓1)getθ=π/4f≡P7,+⇒f(θ)=2Dxf(θ)=0g≡P8,−⇒g(θ)=0Dxg(θ)=32h≡P7,+⇒h(θ)=2Dxh(θ)=0dFdx(θ)=ddx(fgh)(θ)=(f′gh+fg′h+fgh′)(θ)=0+2×32×2+0=128
Commented by mnjuly1970 last updated on 19/Jul/22
very nice mamnon ostad
verynicemamnonostad
Commented by mahdipoor last updated on 20/Jul/22
lotf darid ostad :)
lotfdaridostad:)
Answered by mr W last updated on 20/Jul/22
F(x)=f(x)g(x)h(x) ln F(x)=ln f(x)+ln g(x)+ln h(x) ((F′(x))/(F(x)))=((f′(x))/(f(x)))+((g′(x))/(g(x)))+((h′(x))/(h(x))) F′(x)=g(x)h(x)f′(x)+f(x)h(x)g′(x)+f(x)g(x)h′(x) f((π/4))=h((π/4))=2 g((π/4))=0 F′((π/4))=f((π/4))h((π/4))g′((π/4))=4×g′((π/4)) =4×[((8 tan^7 x)/(cos^2 x))+((8 cot^7 x)/(sin^2 x))]_(x=(π/4)) =32×[(1/(1/2))+(1/(1/2))] =32×4=128
F(x)=f(x)g(x)h(x)lnF(x)=lnf(x)+lng(x)+lnh(x)F′(x)F(x)=f′(x)f(x)+g′(x)g(x)+h′(x)h(x)F′(x)=g(x)h(x)f′(x)+f(x)h(x)g′(x)+f(x)g(x)h′(x)f(π4)=h(π4)=2g(π4)=0F′(π4)=f(π4)h(π4)g′(π4)=4×g′(π4)=4×[8tan7xcos2x+8cot7xsin2x]x=π4=32×[112+112]=32×4=128
Commented by mnjuly1970 last updated on 20/Jul/22
really nice solution master
reallynicesolutionmaster

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