Question Number 117486 by Canovas last updated on 12/Oct/20 | ||
Answered by john santu last updated on 12/Oct/20 | ||
$$\frac{{dy}}{{dx}}\:=\:\mathrm{2}{x}.\left(\mathrm{2}^{{y}} \right)+{x}^{\mathrm{2}} .\mathrm{ln}\:\left(\mathrm{2}\right).\mathrm{2}^{{y}} \:\frac{{dy}}{{dx}} \\ $$$$\left(\mathrm{1}â\mathrm{ln}\:\left(\mathrm{2}\right){x}^{\mathrm{2}} .\mathrm{2}^{{y}} \right)\frac{{dy}}{{dx}}\:=\:\mathrm{2}{x}.\mathrm{2}^{{y}} \\ $$$$\frac{{dy}}{{dx}}\:=\:\frac{{x}.\mathrm{2}^{{y}+\mathrm{1}} }{\mathrm{1}â{y}.\mathrm{ln}\:\left(\mathrm{2}\right)}\:. \\ $$ | ||
Commented by Canovas last updated on 12/Oct/20 | ||
$${Thanks}\:{sir}! \\ $$ | ||