Question Number 162809 by mkam last updated on 01/Jan/22
$${if}\:{y}\:=\:{x}\:+\:\frac{\mathrm{1}}{{x}+\frac{\mathrm{1}}{{x}+\frac{\mathrm{1}}{{x}}._{._{._{._{.} } } } }}\:\:{find}\:{y}^{'} \\ $$
Answered by mindispower last updated on 01/Jan/22
$${y}={x}+\frac{\mathrm{1}}{{y}} \\ $$$$\Rightarrow{y}'=−\frac{{y}'}{{y}^{\mathrm{2}} }+\mathrm{1} \\ $$$${y}'=\frac{{y}^{\mathrm{2}} }{{y}^{\mathrm{2}} +\mathrm{1}} \\ $$
Commented by mkam last updated on 01/Jan/22
$${put}\:{y}^{'} =\:\frac{{y}}{\mathrm{2}{y}−{x}} \\ $$
Answered by MJS_new last updated on 01/Jan/22
$${y}={x}+\frac{\mathrm{1}}{{y}}\:\Rightarrow\:{y}=\frac{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}}\:\Rightarrow\:{y}'=\frac{\mathrm{1}}{\mathrm{2}}+\frac{{x}}{\mathrm{2}\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}} \\ $$