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Question Number 117486 by Canovas last updated on 12/Oct/20

Answered by john santu last updated on 12/Oct/20

(dy/dx) = 2x.(2^y )+x^2 .ln (2).2^y (dy/dx) (1−ln (2)x^2 .2^y )(dy/dx) = 2x.2^y (dy/dx) = ((x.2^(y+1) )/(1−y.ln (2))) .

$$\frac{{dy}}{{dx}}\:=\:\mathrm{2}{x}.\left(\mathrm{2}^{{y}} \right)+{x}^{\mathrm{2}} .\mathrm{ln}\:\left(\mathrm{2}\right).\mathrm{2}^{{y}} \:\frac{{dy}}{{dx}} \\ $$$$\left(\mathrm{1}−\mathrm{ln}\:\left(\mathrm{2}\right){x}^{\mathrm{2}} .\mathrm{2}^{{y}} \right)\frac{{dy}}{{dx}}\:=\:\mathrm{2}{x}.\mathrm{2}^{{y}} \\ $$$$\frac{{dy}}{{dx}}\:=\:\frac{{x}.\mathrm{2}^{{y}+\mathrm{1}} }{\mathrm{1}−{y}.\mathrm{ln}\:\left(\mathrm{2}\right)}\:. \\ $$

Commented by Canovas last updated on 12/Oct/20

Thanks sir!

$${Thanks}\:{sir}! \\ $$

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