Menu Close

let-f-x-determinant-x-x-2-x-3-0-2x-3x-2-1-0-x-find-f-x-




Question Number 138167 by physicstutes last updated on 10/Apr/21
let f(x) =  determinant ((x,x^2 ,x^3 ),(0,(2x ),(3x^2 )),(1,0,x)) find   f ′(x)
$$\mathrm{let}\:{f}\left({x}\right)\:=\:\begin{vmatrix}{{x}}&{{x}^{\mathrm{2}} }&{{x}^{\mathrm{3}} }\\{\mathrm{0}}&{\mathrm{2}{x}\:}&{\mathrm{3}{x}^{\mathrm{2}} }\\{\mathrm{1}}&{\mathrm{0}}&{{x}}\end{vmatrix}\:\mathrm{find} \\ $$$$\:{f}\:'\left({x}\right)\: \\ $$
Answered by Dwaipayan Shikari last updated on 10/Apr/21
f(x)=x(2x^2 )−x^2 (−3x^2 )+x^3 (0−2x)  =2x^3 +3x^4 −2x^4 =2x^3 +x^4   f′(x)=6x^2 +4x^3
$${f}\left({x}\right)={x}\left(\mathrm{2}{x}^{\mathrm{2}} \right)−{x}^{\mathrm{2}} \left(−\mathrm{3}{x}^{\mathrm{2}} \right)+{x}^{\mathrm{3}} \left(\mathrm{0}−\mathrm{2}{x}\right) \\ $$$$=\mathrm{2}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{4}} =\mathrm{2}{x}^{\mathrm{3}} +{x}^{\mathrm{4}} \\ $$$${f}'\left({x}\right)=\mathrm{6}{x}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{3}} \\ $$
Answered by mathmax by abdo last updated on 10/Apr/21
f(x)=x determinant (((2x        3x^2 )),((0               x)))−o+1 determinant (((x^2         x^3 )),((2x        3x^2 )))  =x(2x^2 )+(3x^4 −2x^4 ) =x^4  +2x^3  ⇒f^′ (x)=4x^3  +6x^2
$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}\begin{vmatrix}{\mathrm{2x}\:\:\:\:\:\:\:\:\mathrm{3x}^{\mathrm{2}} }\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}}\end{vmatrix}−\mathrm{o}+\mathrm{1}\begin{vmatrix}{\mathrm{x}^{\mathrm{2}} \:\:\:\:\:\:\:\:\mathrm{x}^{\mathrm{3}} }\\{\mathrm{2x}\:\:\:\:\:\:\:\:\mathrm{3x}^{\mathrm{2}} }\end{vmatrix} \\ $$$$=\mathrm{x}\left(\mathrm{2x}^{\mathrm{2}} \right)+\left(\mathrm{3x}^{\mathrm{4}} −\mathrm{2x}^{\mathrm{4}} \right)\:=\mathrm{x}^{\mathrm{4}} \:+\mathrm{2x}^{\mathrm{3}} \:\Rightarrow\mathrm{f}^{'} \left(\mathrm{x}\right)=\mathrm{4x}^{\mathrm{3}} \:+\mathrm{6x}^{\mathrm{2}} \\ $$

Leave a Reply

Your email address will not be published.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