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Question Number 106847 by mathocean1 last updated on 07/Aug/20

Given f(x)=((3(√3))/(sinx))+(1/(cosx)) show that f ′(x)=cosx(((tan^3 x−3(√3)))/(sin^2 x))

$${Given}\:{f}\left({x}\right)=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{{sinx}}+\frac{\mathrm{1}}{{cosx}} \\ $$$${show}\:{that}\:{f}\:'\left({x}\right)={cosx}\frac{\left({tan}^{\mathrm{3}} {x}−\mathrm{3}\sqrt{\mathrm{3}}\right)}{{sin}^{\mathrm{2}} {x}} \\ $$

Answered by bemath last updated on 07/Aug/20

^(@bemath@) f(x)=3(√3) csc x + sec x f ′(x)=−3(√3) cot x csc x+sec x tan x f ′(x) = ((sin x)/(cos^2 x))−((3(√3) cos x)/(sin^2 x)) = ((sin^3 x−3(√3) cos^3 x)/(cos^2 x sin^2 x)) =((tan^3 x−3(√3) )/(sin^2 x((1/(cos x))))) = ((cos x(tan^3 x−3(√3)))/(sin^2 x))

$$\:\:\:\:\:^{@\mathrm{bemath}@} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{3}\sqrt{\mathrm{3}}\:\mathrm{csc}\:\mathrm{x}\:+\:\mathrm{sec}\:\mathrm{x}\: \\ $$$$\mathrm{f}\:'\left(\mathrm{x}\right)=−\mathrm{3}\sqrt{\mathrm{3}}\:\mathrm{cot}\:\mathrm{x}\:\mathrm{csc}\:\mathrm{x}+\mathrm{sec}\:\mathrm{x}\:\mathrm{tan}\:\mathrm{x} \\ $$$$\mathrm{f}\:'\left(\mathrm{x}\right)\:=\:\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}−\frac{\mathrm{3}\sqrt{\mathrm{3}}\:\mathrm{cos}\:\mathrm{x}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}} \\ $$$$=\:\frac{\mathrm{sin}\:^{\mathrm{3}} \mathrm{x}−\mathrm{3}\sqrt{\mathrm{3}}\:\mathrm{cos}\:^{\mathrm{3}} \mathrm{x}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\:\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}} \\ $$$$=\frac{\mathrm{tan}\:^{\mathrm{3}} \mathrm{x}−\mathrm{3}\sqrt{\mathrm{3}}\:}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\left(\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{x}}\right)}\:=\:\frac{\mathrm{cos}\:\mathrm{x}\left(\mathrm{tan}\:^{\mathrm{3}} \mathrm{x}−\mathrm{3}\sqrt{\mathrm{3}}\right)}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}\: \\ $$

Commented by mathocean1 last updated on 07/Aug/20

Please sir can you detail the last line? I′m not understanding how tan^(3 ) appeared...

$${Please}\:{sir}\:{can}\:{you}\:{detail}\:{the}\:{last} \\ $$$${line}?\:{I}'{m}\:{not}\:{understanding}\:{how} \\ $$$${tan}^{\mathrm{3}\:} \:{appeared}... \\ $$

Commented by bemath last updated on 07/Aug/20

from ((sin^3 x−3(√3) cos^3 x)/(cos^2 x sin^2 x)) :((cos^3 x)/(cos^3 x))

$$\mathrm{from}\:\frac{\mathrm{sin}\:^{\mathrm{3}} \mathrm{x}−\mathrm{3}\sqrt{\mathrm{3}}\:\mathrm{cos}\:^{\mathrm{3}} \mathrm{x}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\:\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}\::\frac{\mathrm{cos}\:^{\mathrm{3}} \mathrm{x}}{\mathrm{cos}\:^{\mathrm{3}} \mathrm{x}}\: \\ $$

Answered by Dwaipayan Shikari last updated on 07/Aug/20

f(x)=((3(√3))/(sinx))+(1/(cosx)) f′(x)=((−3(√3)cosx)/(sin^2 x))+secxtanx=((−3(√(3 ))cosx)/(sin^2 x))+((sinx)/(cos^2 x)) =((−3(√3)cos^3 x+sin^3 x)/(sin^2 xcos^2 x))=((−3(√3)+((sin^3 x)/(cos^3 x)))/(sin^2 x((cos^2 x)/(cos^3 x))))=cosx(((tan^3 x−3(√3))/(sin^2 x)))

$${f}\left({x}\right)=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{{sinx}}+\frac{\mathrm{1}}{{cosx}} \\ $$$${f}'\left({x}\right)=\frac{−\mathrm{3}\sqrt{\mathrm{3}}{cosx}}{{sin}^{\mathrm{2}} {x}}+{secxtanx}=\frac{−\mathrm{3}\sqrt{\mathrm{3}\:}{cosx}}{{sin}^{\mathrm{2}} {x}}+\frac{{sinx}}{{cos}^{\mathrm{2}} {x}} \\ $$$$=\frac{−\mathrm{3}\sqrt{\mathrm{3}}{cos}^{\mathrm{3}} {x}+{sin}^{\mathrm{3}} {x}}{{sin}^{\mathrm{2}} {xcos}^{\mathrm{2}} {x}}=\frac{−\mathrm{3}\sqrt{\mathrm{3}}+\frac{{sin}^{\mathrm{3}} {x}}{{cos}^{\mathrm{3}} {x}}}{{sin}^{\mathrm{2}} {x}\frac{{cos}^{\mathrm{2}} {x}}{{cos}^{\mathrm{3}} {x}}}={cosx}\left(\frac{{tan}^{\mathrm{3}} {x}−\mathrm{3}\sqrt{\mathrm{3}}}{{sin}^{\mathrm{2}} {x}}\right) \\ $$

Commented by bemath last updated on 07/Aug/20

typo sir. it tan^3 x

$$\mathrm{typo}\:\mathrm{sir}.\:\mathrm{it}\:\mathrm{tan}\:^{\mathrm{3}} \mathrm{x}\: \\ $$

Answered by 1549442205PVT last updated on 07/Aug/20

f(x)=((3(√3))/(sinx))+(1/(cosx)) ⇒f ′(x)=−((3(√3))/(sin^2 x))×cosx+((−1)/(cos^2 x))×(−sinx) (since ((1/u))′=((−1)/u^2 )×u′) =((−3(√3)cosx)/(sin^2 x))+((sinx)/(cos^2 x))=((−3(√3)cosx)/(sin^2 x))+((sin^3 x)/(sin^2 xcos^2 x)) =((−3(√3)cosx)/(sin^2 x))+(((((sin^3 x)/(cos^3 x)))cosx)/(sin^2 x)) =((−3(√3)cosx)/(sin^2 x))+((tan^3 x.cosx)/(sin^2 x)) =cosx(((tan^3 x−3(√3))/(sin^2 x)))(q.e.d)

$${f}\left({x}\right)=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{{sinx}}+\frac{\mathrm{1}}{{cosx}} \\ $$$$\Rightarrow\mathrm{f}\:'\left(\mathrm{x}\right)=−\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}×\mathrm{cosx}+\frac{−\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \mathrm{x}}×\left(−\mathrm{sinx}\right) \\ $$$$\left(\mathrm{since}\:\left(\frac{\mathrm{1}}{\mathrm{u}}\right)'=\frac{−\mathrm{1}}{\mathrm{u}^{\mathrm{2}} }×\mathrm{u}'\right) \\ $$$$=\frac{−\mathrm{3}\sqrt{\mathrm{3}}\mathrm{cosx}}{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}+\frac{\mathrm{sinx}}{\mathrm{cos}^{\mathrm{2}} \mathrm{x}}=\frac{−\mathrm{3}\sqrt{\mathrm{3}}\mathrm{cosx}}{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}+\frac{\mathrm{sin}^{\mathrm{3}} \mathrm{x}}{\mathrm{sin}^{\mathrm{2}} \mathrm{xcos}^{\mathrm{2}} \mathrm{x}} \\ $$$$=\frac{−\mathrm{3}\sqrt{\mathrm{3}}\mathrm{cosx}}{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}+\frac{\left(\frac{\mathrm{sin}^{\mathrm{3}} \mathrm{x}}{\mathrm{cos}^{\mathrm{3}} \mathrm{x}}\right)\mathrm{cosx}}{\mathrm{sin}^{\mathrm{2}} \mathrm{x}} \\ $$$$=\frac{−\mathrm{3}\sqrt{\mathrm{3}}\mathrm{cosx}}{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}+\frac{\mathrm{tan}^{\mathrm{3}} \mathrm{x}.\mathrm{cosx}}{\mathrm{sin}^{\mathrm{2}} \mathrm{x}} \\ $$$$=\boldsymbol{\mathrm{cosx}}\left(\frac{\boldsymbol{\mathrm{tan}}^{\mathrm{3}} \boldsymbol{\mathrm{x}}−\mathrm{3}\sqrt{\mathrm{3}}}{\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}}\right)\left(\boldsymbol{\mathrm{q}}.\boldsymbol{\mathrm{e}}.\boldsymbol{\mathrm{d}}\right) \\ $$

Commented by mathocean1 last updated on 08/Aug/20

Thank you sirs...

$${Thank}\:{you}\:{sirs}... \\ $$

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