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Question Number 118969 by benjo_mathlover last updated on 21/Oct/20
Given f(x) = ((sin x+cos x)/(sin x−cos x)) find the value of f ′′(x) + f ′(x) + 1 .
$$\:{Given}\:{f}\left({x}\right)\:=\:\frac{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}−\mathrm{cos}\:{x}} \\ $$$${find}\:{the}\:{value}\:{of}\: \\ $$$$\:{f}\:''\left({x}\right)\:+\:{f}\:'\left({x}\right)\:+\:\mathrm{1}\:. \\ $$
Answered by TANMAY PANACEA last updated on 21/Oct/20
f(x)=((tanx+1)/(tanx−1))=−tan((π/4)+x) f^′ (x)=−sec^2 ((π/4)+x) f^(′′) (x)=−2sec^2 ((π/4)+x)tan((π/4)+x) 1+f^′ (x)+f^(′′) (x)=1−2sec^2 ((π/4)+x)tan((π/4)+x)−sec^2 ((π/4)+x)
$${f}\left({x}\right)=\frac{{tanx}+\mathrm{1}}{{tanx}−\mathrm{1}}=−{tan}\left(\frac{\pi}{\mathrm{4}}+{x}\right) \\ $$$${f}^{'} \left({x}\right)=−{sec}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}+{x}\right) \\ $$$${f}^{''} \left({x}\right)=−\mathrm{2}{sec}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}+{x}\right){tan}\left(\frac{\pi}{\mathrm{4}}+{x}\right) \\ $$$$\mathrm{1}+{f}^{'} \left({x}\right)+{f}^{''} \left({x}\right)=\mathrm{1}−\mathrm{2}{sec}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}+{x}\right){tan}\left(\frac{\pi}{\mathrm{4}}+{x}\right)−{sec}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}+{x}\right) \\ $$$$ \\ $$
Answered by bramlexs22 last updated on 21/Oct/20
f ′(x) = (((cos x−sin x)(sin x−cos x)−(sin x+cos x)(cos x+sin x))/(1−sin 2x)) = ((sin 2x−1−(1+sin 2x))/(1−sin 2x)) = ((−2)/(1−sin 2x)) f ′′(x) = ((0−(−2)(−2cos 2x))/((1−sin 2x)^2 )) = ((−4cos 2x)/((1−sin 2x)^2 )) f ′′(x)+f ′(x)+1 = ((−4cos 2x−2(1−sin 2x)+(1−sin 2x)^2 )/((1−sin 2x)^2 )) = ((−4cos 2x+2sin 2x−2+1−2sin 2x+sin^2 2x)/((1−sin 2x)^2 )) = ((sin^2 2x−4cos 2x−1)/((1−sin 2x)^2 )) = ((1−cos^2 2x−4cos 2x−1)/((1−sin 2x)^2 )) = ((−cos 2x(cos 2x+4))/((1−sin 2x)^2 ))
$$\:{f}\:'\left({x}\right)\:=\:\frac{\left(\mathrm{cos}\:{x}−\mathrm{sin}\:{x}\right)\left(\mathrm{sin}\:{x}−\mathrm{cos}\:{x}\right)−\left(\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right)\left(\mathrm{cos}\:{x}+\mathrm{sin}\:{x}\right)}{\mathrm{1}−\mathrm{sin}\:\mathrm{2}{x}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{sin}\:\mathrm{2}{x}−\mathrm{1}−\left(\mathrm{1}+\mathrm{sin}\:\mathrm{2}{x}\right)}{\mathrm{1}−\mathrm{sin}\:\mathrm{2}{x}}\:=\:\frac{−\mathrm{2}}{\mathrm{1}−\mathrm{sin}\:\mathrm{2}{x}} \\ $$$${f}\:''\left({x}\right)\:=\:\frac{\mathrm{0}−\left(−\mathrm{2}\right)\left(−\mathrm{2cos}\:\mathrm{2}{x}\right)}{\left(\mathrm{1}−\mathrm{sin}\:\mathrm{2}{x}\right)^{\mathrm{2}} }\:=\:\frac{−\mathrm{4cos}\:\mathrm{2}{x}}{\left(\mathrm{1}−\mathrm{sin}\:\mathrm{2}{x}\right)^{\mathrm{2}} } \\ $$$$ \\ $$$${f}\:''\left({x}\right)+{f}\:'\left({x}\right)+\mathrm{1}\:=\:\frac{−\mathrm{4cos}\:\mathrm{2}{x}−\mathrm{2}\left(\mathrm{1}−\mathrm{sin}\:\mathrm{2}{x}\right)+\left(\mathrm{1}−\mathrm{sin}\:\mathrm{2}{x}\right)^{\mathrm{2}} }{\left(\mathrm{1}−\mathrm{sin}\:\mathrm{2}{x}\right)^{\mathrm{2}} } \\ $$$$\:\:=\:\frac{−\mathrm{4cos}\:\mathrm{2}{x}+\mathrm{2sin}\:\mathrm{2}{x}−\mathrm{2}+\mathrm{1}−\mathrm{2sin}\:\mathrm{2}{x}+\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}{x}}{\left(\mathrm{1}−\mathrm{sin}\:\mathrm{2}{x}\right)^{\mathrm{2}} } \\ $$$$\:\:=\:\frac{\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}{x}−\mathrm{4cos}\:\mathrm{2}{x}−\mathrm{1}}{\left(\mathrm{1}−\mathrm{sin}\:\mathrm{2}{x}\right)^{\mathrm{2}} }\:=\:\frac{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{2}{x}−\mathrm{4cos}\:\mathrm{2}{x}−\mathrm{1}}{\left(\mathrm{1}−\mathrm{sin}\:\mathrm{2}{x}\right)^{\mathrm{2}} } \\ $$$$\:=\:\frac{−\mathrm{cos}\:\mathrm{2}{x}\left(\mathrm{cos}\:\mathrm{2}{x}+\mathrm{4}\right)}{\left(\mathrm{1}−\mathrm{sin}\:\mathrm{2}{x}\right)^{\mathrm{2}} }\: \\ $$
Answered by MJS_new last updated on 21/Oct/20
sin x =((tan x)/( (√(1+tan^2 x))))∧cos x =(1/( (√(1+tan^2 x)))) let t=tan x ⇒ t′=1+t^2 ⇒ f(x)=((t+1)/(t−1)) f′(x)=−((2t′)/((t−1)^2 ))=−((2(t^2 +1))/((t−1)^2 )) f′′(x)=((4t′(t+1))/((t−1)^3 ))=((4(t+1)(t^2 +1))/((t−1)^3 )) 1+f′(x)+f′′(x)=(((t+1)(3t^2 +5))/((t−1)^3 ))= =−(((1+tan x)(5+3tan^2 x))/((1−tan x)^3 ))= =(((sin x +cos x)(3+2cos^2 x))/((sin x −cos x)^3 ))
$$\mathrm{sin}\:{x}\:=\frac{\mathrm{tan}\:{x}}{\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{x}}}\wedge\mathrm{cos}\:{x}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{x}}} \\ $$$$\mathrm{let}\:{t}=\mathrm{tan}\:{x}\:\Rightarrow\:{t}'=\mathrm{1}+{t}^{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$${f}\left({x}\right)=\frac{{t}+\mathrm{1}}{{t}−\mathrm{1}} \\ $$$${f}'\left({x}\right)=−\frac{\mathrm{2}{t}'}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }=−\frac{\mathrm{2}\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${f}''\left({x}\right)=\frac{\mathrm{4}{t}'\left({t}+\mathrm{1}\right)}{\left({t}−\mathrm{1}\right)^{\mathrm{3}} }=\frac{\mathrm{4}\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{\left({t}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\mathrm{1}+{f}'\left({x}\right)+{f}''\left({x}\right)=\frac{\left({t}+\mathrm{1}\right)\left(\mathrm{3}{t}^{\mathrm{2}} +\mathrm{5}\right)}{\left({t}−\mathrm{1}\right)^{\mathrm{3}} }= \\ $$$$=−\frac{\left(\mathrm{1}+\mathrm{tan}\:{x}\right)\left(\mathrm{5}+\mathrm{3tan}^{\mathrm{2}} \:{x}\right)}{\left(\mathrm{1}−\mathrm{tan}\:{x}\right)^{\mathrm{3}} }= \\ $$$$=\frac{\left(\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}\right)\left(\mathrm{3}+\mathrm{2cos}^{\mathrm{2}} \:{x}\right)}{\left(\mathrm{sin}\:{x}\:−\mathrm{cos}\:{x}\right)^{\mathrm{3}} } \\ $$

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