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Question Number 120912 by bemath last updated on 04/Nov/20

find (dy/dx) from equation ((xy^3 )/(sec y−3)) = 1+y^4

$$\:\mathrm{find}\:\frac{\mathrm{dy}}{\mathrm{dx}}\:\mathrm{from}\:\mathrm{equation}\: \\ $$$$\:\frac{\mathrm{xy}^{\mathrm{3}} }{\mathrm{sec}\:\mathrm{y}−\mathrm{3}}\:=\:\mathrm{1}+\mathrm{y}^{\mathrm{4}} \\ $$

Answered by liberty last updated on 04/Nov/20

by implicit differential ⇒ ((xy^3 )/(1+y^4 )) = sec y−3 ⇒ (((y^3 +3xy^2 y′)(1+y^4 )−(4y^3 y′)(xy^3 ))/((1+y^4 )^2 )) = sec y tan y .y′ ⇒ ((y^3 +y^7 +(3xy^2 +3xy^6 )y′−4xy^6 y′)/((1+y^4 )^2 )) = sec y tan y. y′ ⇒y^3 +y^7 +(3xy^2 −xy^6 )y′=(1+y^4 )^2 sec y tan y. y′ ⇒y′ = ((y^3 +y^7 )/({(1+y^4 )^2 sec y tan y+xy^6 −3xy^2 })).▲

$$\mathrm{by}\:\mathrm{implicit}\:\mathrm{differential}\: \\ $$$$\Rightarrow\:\frac{\mathrm{xy}^{\mathrm{3}} }{\mathrm{1}+\mathrm{y}^{\mathrm{4}} }\:=\:\mathrm{sec}\:\mathrm{y}−\mathrm{3} \\ $$$$\Rightarrow\:\frac{\left(\mathrm{y}^{\mathrm{3}} +\mathrm{3xy}^{\mathrm{2}} \mathrm{y}'\right)\left(\mathrm{1}+\mathrm{y}^{\mathrm{4}} \right)−\left(\mathrm{4y}^{\mathrm{3}} \mathrm{y}'\right)\left(\mathrm{xy}^{\mathrm{3}} \right)}{\left(\mathrm{1}+\mathrm{y}^{\mathrm{4}} \right)^{\mathrm{2}} }\:=\:\mathrm{sec}\:\mathrm{y}\:\mathrm{tan}\:\mathrm{y}\:.\mathrm{y}' \\ $$$$\Rightarrow\:\frac{\mathrm{y}^{\mathrm{3}} +\mathrm{y}^{\mathrm{7}} +\left(\mathrm{3xy}^{\mathrm{2}} +\mathrm{3xy}^{\mathrm{6}} \right)\mathrm{y}'−\mathrm{4xy}^{\mathrm{6}} \mathrm{y}'}{\left(\mathrm{1}+\mathrm{y}^{\mathrm{4}} \right)^{\mathrm{2}} }\:=\:\mathrm{sec}\:\mathrm{y}\:\mathrm{tan}\:\mathrm{y}.\:\mathrm{y}' \\ $$$$\Rightarrow\mathrm{y}^{\mathrm{3}} +\mathrm{y}^{\mathrm{7}} +\left(\mathrm{3xy}^{\mathrm{2}} −\mathrm{xy}^{\mathrm{6}} \right)\mathrm{y}'=\left(\mathrm{1}+\mathrm{y}^{\mathrm{4}} \right)^{\mathrm{2}} \mathrm{sec}\:\mathrm{y}\:\mathrm{tan}\:\mathrm{y}.\:\mathrm{y}' \\ $$$$\Rightarrow\mathrm{y}'\:=\:\frac{\mathrm{y}^{\mathrm{3}} +\mathrm{y}^{\mathrm{7}} }{\left\{\left(\mathrm{1}+\mathrm{y}^{\mathrm{4}} \right)^{\mathrm{2}} \mathrm{sec}\:\mathrm{y}\:\mathrm{tan}\:\mathrm{y}+\mathrm{xy}^{\mathrm{6}} −\mathrm{3xy}^{\mathrm{2}} \right\}}.\blacktriangle \\ $$

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