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Question Number 66101 by Rio Michael last updated on 09/Aug/19

find (dy/dx)  when y = x^2 ln(3x)  Given that xsinx − y^2 =0 show that  y^2  = 2cosx −2((dy/dx))^2  −2y(d^2 y/dx^2 )

$${find}\:\frac{{dy}}{{dx}}\:\:{when}\:{y}\:=\:{x}^{\mathrm{2}} {ln}\left(\mathrm{3}{x}\right) \\ $$$${Given}\:{that}\:{xsinx}\:−\:{y}^{\mathrm{2}} =\mathrm{0}\:{show}\:{that}\:\:{y}^{\mathrm{2}} \:=\:\mathrm{2}{cosx}\:−\mathrm{2}\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} \:−\mathrm{2}{y}\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} } \\ $$

Commented by Prithwish sen last updated on 09/Aug/19

(dy/dx)=2xln3x +x  y^2 =xsinx  2y(dy/dx)=sinx+xcosx⇒(dy/dx) = ((sinx+xcosx)/(2(√(xsinx))))  (d^2 y/dx^2 ) =(1/2). ((y(2cosx−xsinx)−(((sinx+xcosx)^2 )/(2y)))/y^2 )   4y^3  (d^2 y/dx^2 ) = 2y^2 (2cosx−y^2 )−(2y(dy/dx))^2   2y(d^2 y/dx^2 ) = 2cosx −y^2  −2((dy/dx))^2   proved.

$$\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{2xln3x}\:+\mathrm{x} \\ $$$$\mathrm{y}^{\mathrm{2}} =\mathrm{xsinx} \\ $$$$\mathrm{2y}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{sinx}+\mathrm{xcosx}\Rightarrow\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{sinx}+\mathrm{xcosx}}{\mathrm{2}\sqrt{\mathrm{xsinx}}} \\ $$$$\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{2}}.\:\frac{\mathrm{y}\left(\mathrm{2cosx}−\mathrm{xsinx}\right)−\frac{\left(\mathrm{sinx}+\mathrm{xcosx}\right)^{\mathrm{2}} }{\mathrm{2y}}}{\mathrm{y}^{\mathrm{2}} } \\ $$$$\:\mathrm{4y}^{\mathrm{3}} \:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:=\:\mathrm{2y}^{\mathrm{2}} \left(\mathrm{2cosx}−\mathrm{y}^{\mathrm{2}} \right)−\left(\mathrm{2y}\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{\mathrm{2}} \\ $$$$\mathrm{2y}\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:=\:\mathrm{2cosx}\:−\mathrm{y}^{\mathrm{2}} \:−\mathrm{2}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{\mathrm{2}} \\ $$$$\mathrm{proved}. \\ $$

Answered by GordonYeeman last updated on 09/Aug/19

(dy/dx)=2xln(3x)+x^2 (1/(3x))3=x(2ln(3x)+1)

$$\frac{{dy}}{{dx}}=\mathrm{2}{xln}\left(\mathrm{3}{x}\right)+{x}^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{3}{x}}\mathrm{3}={x}\left(\mathrm{2}{ln}\left(\mathrm{3}{x}\right)+\mathrm{1}\right) \\ $$

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