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Question Number 64350 by Rio Michael last updated on 16/Jul/19

Given that f(x) =  determinant ((x,x^2 ,x^3 ),(1,(2x),(3x^2 )),(0,2,(6x))), find f ′ (x)

$${Given}\:{that}\:{f}\left({x}\right)\:=\:\begin{vmatrix}{{x}}&{{x}^{\mathrm{2}} }&{{x}^{\mathrm{3}} }\\{\mathrm{1}}&{\mathrm{2}{x}}&{\mathrm{3}{x}^{\mathrm{2}} }\\{\mathrm{0}}&{\mathrm{2}}&{\mathrm{6}{x}}\end{vmatrix},\:{find}\:{f}\:'\:\left({x}\right) \\ $$

Commented by Tony Lin last updated on 17/Jul/19

f(x)=12x^3 +2x^3 −6x^3 −6x^3 =2x^3   f′(x)=6x^2

$${f}\left({x}\right)=\mathrm{12}{x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{3}} −\mathrm{6}{x}^{\mathrm{3}} −\mathrm{6}{x}^{\mathrm{3}} =\mathrm{2}{x}^{\mathrm{3}} \\ $$$${f}'\left({x}\right)=\mathrm{6}{x}^{\mathrm{2}} \\ $$

Commented by Rio Michael last updated on 17/Jul/19

thanks

$${thanks} \\ $$

Commented by Prithwish sen last updated on 17/Jul/19

f′(x)= determinant (((1   2x   3x^2 )),((0    2      6x))) = 12x^2 −6x^2  = 6x^2                     0     0      6

$$\mathrm{f}'\left(\mathrm{x}\right)=\begin{vmatrix}{\mathrm{1}\:\:\:\mathrm{2x}\:\:\:\mathrm{3x}^{\mathrm{2}} }\\{\mathrm{0}\:\:\:\:\mathrm{2}\:\:\:\:\:\:\mathrm{6x}}\end{vmatrix}\:=\:\mathrm{12x}^{\mathrm{2}} −\mathrm{6x}^{\mathrm{2}} \:=\:\mathrm{6x}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\mathrm{6} \\ $$

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