Menu Close

Given-f-x-7-cos-2x-2sin-2-x-find-f-10-x-




Question Number 114066 by john santu last updated on 17/Sep/20
Given f(x)=7+cos 2x+2sin^2 x  find f^((10)) (x) ?
$${Given}\:{f}\left({x}\right)=\mathrm{7}+\mathrm{cos}\:\mathrm{2}{x}+\mathrm{2sin}\:^{\mathrm{2}} {x} \\ $$$${find}\:{f}^{\left(\mathrm{10}\right)} \left({x}\right)\:? \\ $$
Commented by bemath last updated on 17/Sep/20
f(x)=7+1−2sin^2 x+2sin^2 x  f(x)=8 ⇒f^((10)) (x)= 0
$${f}\left({x}\right)=\mathrm{7}+\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} {x}+\mathrm{2sin}\:^{\mathrm{2}} {x} \\ $$$${f}\left({x}\right)=\mathrm{8}\:\Rightarrow{f}\:^{\left(\mathrm{10}\right)} \left({x}\right)=\:\mathrm{0}\: \\ $$
Answered by Olaf last updated on 17/Sep/20
f(x) = 7+cos2x+2sin^2 x  f(x) = 7+cos2x+2((1−cos2x)/2)  f(x) = 8  f^((k)) (x) = 0
$${f}\left({x}\right)\:=\:\mathrm{7}+\mathrm{cos2}{x}+\mathrm{2sin}^{\mathrm{2}} {x} \\ $$$${f}\left({x}\right)\:=\:\mathrm{7}+\mathrm{cos2}{x}+\mathrm{2}\frac{\mathrm{1}−\mathrm{cos2}{x}}{\mathrm{2}} \\ $$$${f}\left({x}\right)\:=\:\mathrm{8} \\ $$$${f}^{\left({k}\right)} \left({x}\right)\:=\:\mathrm{0} \\ $$
Answered by 1549442205PVT last updated on 17/Sep/20
Since cos2x=1−2sin^2 x,cos2x+2sin^2 x  =1⇒f(x)=7+cos2x+2sin^2 x=8  ⇒f ′(x)=0⇒f^((n)) (x)=0∀n≥1,n∈N
$$\mathrm{Since}\:\mathrm{cos2x}=\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \mathrm{x},\mathrm{cos2x}+\mathrm{2sin}^{\mathrm{2}} \mathrm{x} \\ $$$$=\mathrm{1}\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\mathrm{7}+\mathrm{cos2x}+\mathrm{2sin}^{\mathrm{2}} \mathrm{x}=\mathrm{8} \\ $$$$\Rightarrow\mathrm{f}\:'\left(\mathrm{x}\right)=\mathrm{0}\Rightarrow\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{x}\right)=\mathrm{0}\forall\mathrm{n}\geqslant\mathrm{1},\mathrm{n}\in\mathrm{N} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *