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Question Number 169303 by Mastermind last updated on 28/Apr/22
Given that:  x cos y=sin(x+y), find (dy/dx)    Mastermind
$${Given}\:{that}: \\ $$$${x}\:{cos}\:{y}={sin}\left({x}+{y}\right),\:{find}\:\frac{{dy}}{{dx}} \\ $$$$ \\ $$$${Mastermind} \\ $$
Commented by infinityaction last updated on 28/Apr/22
x = ((sin xcos y+sin ycos x)/(cos y))  x = sin x + tan ycos x  xsec x −tan x = tan y  D.W.R.T.X  sec^2 y (dy/dx) = x sec xtan x + sec x− sec^2 x  (dy/(dx )) = sec x(((xtan x−sec x+1)/(sec^2 y)))
$${x}\:=\:\frac{\mathrm{sin}\:{x}\mathrm{cos}\:{y}+\mathrm{sin}\:{y}\mathrm{cos}\:{x}}{\mathrm{cos}\:{y}} \\ $$$${x}\:=\:\mathrm{sin}\:{x}\:+\:\mathrm{tan}\:{y}\mathrm{cos}\:{x} \\ $$$${x}\mathrm{sec}\:{x}\:−\mathrm{tan}\:{x}\:=\:\mathrm{tan}\:{y} \\ $$$${D}.{W}.{R}.{T}.{X} \\ $$$$\mathrm{sec}\:^{\mathrm{2}} {y}\:\frac{{dy}}{{dx}}\:=\:{x}\:\mathrm{sec}\:{x}\mathrm{tan}\:{x}\:+\:\mathrm{sec}\:{x}−\:\mathrm{sec}\:^{\mathrm{2}} {x} \\ $$$$\frac{{dy}}{{dx}\:}\:=\:\mathrm{sec}\:{x}\left(\frac{{x}\mathrm{tan}\:{x}−\mathrm{sec}\:{x}+\mathrm{1}}{\mathrm{sec}\:^{\mathrm{2}} {y}}\right) \\ $$$$ \\ $$$$ \\ $$
Answered by mr W last updated on 28/Apr/22
cos y−x sin y (dy/dx)=cos (x+y)(1+ (dy/dx))  cos y−cos (x+y)=[x sin y+cos (x+y)] (dy/dx)  (dy/dx)=((cos y−cos (x+y))/(x sin y+cos (x+y)))
$$\mathrm{cos}\:{y}−{x}\:\mathrm{sin}\:{y}\:\frac{{dy}}{{dx}}=\mathrm{cos}\:\left({x}+{y}\right)\left(\mathrm{1}+\:\frac{{dy}}{{dx}}\right) \\ $$$$\mathrm{cos}\:{y}−\mathrm{cos}\:\left({x}+{y}\right)=\left[{x}\:\mathrm{sin}\:{y}+\mathrm{cos}\:\left({x}+{y}\right)\right]\:\frac{{dy}}{{dx}} \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{cos}\:{y}−\mathrm{cos}\:\left({x}+{y}\right)}{{x}\:\mathrm{sin}\:{y}+\mathrm{cos}\:\left({x}+{y}\right)} \\ $$

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