Question Number 38252 by Rio Mike last updated on 23/Jun/18
$${if}\:{x}^{\mathrm{2}} \:+\:\mathrm{3}{xy}\:−\:{y}^{\mathrm{2}} \:=\:\mathrm{3}\:{find}\: \\ $$$$\frac{{dy}}{{dx}}\:{at}\:{point}\:\left(\mathrm{1},\mathrm{1}\right)\:{hence} \\ $$$${differentiate}\:\frac{{sin}\:{x}}{\mathrm{1}\:+\:{x}}\:{with}\:{respect} \\ $$$${to}\:{x}. \\ $$
Commented by math khazana by abdo last updated on 23/Jun/18
$${x}^{\mathrm{2}} \:+\mathrm{3}{xy}\:−{y}^{\mathrm{2}} =\mathrm{3}\:\Rightarrow{y}^{\mathrm{2}} \:−\mathrm{3}{xy}\:−{x}^{\mathrm{2}} \:+\mathrm{3}=\mathrm{0} \\ $$$$\Delta=\mathrm{9}{x}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{3}−{x}^{\mathrm{2}} \right)=\mathrm{13}{x}^{\mathrm{2}} \:−\mathrm{12}\:{so}\:{if}\:{x}^{\mathrm{2}} \geqslant\frac{\mathrm{12}}{\mathrm{13}} \\ $$$${y}\left({x}\right)\:=\frac{\mathrm{3}{x}\:\overset{−} {+}\:\sqrt{\mathrm{13}{x}^{\mathrm{2}} −\mathrm{12}}}{\mathrm{2}}\:\Rightarrow \\ $$$${y}^{'} \left({x}\right)=\:\frac{\mathrm{3}}{\mathrm{2}}\:\overset{−} {+}\:\frac{\mathrm{1}}{\mathrm{2}}\:\:\frac{\mathrm{26}{x}}{\mathrm{2}\sqrt{\mathrm{13}{x}^{\mathrm{2}} −\mathrm{12}}}\: \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\:\overset{−} {+}\:\:\frac{\mathrm{13}}{\mathrm{2}\sqrt{\mathrm{13}^{} {x}^{\mathrm{2}} \:−\mathrm{12}}}\:\Rightarrow \\ $$$${y}^{'} \left(\mathrm{1}\right)=\frac{\mathrm{3}}{\mathrm{2}}\:\overset{−} {+}\:\frac{\mathrm{13}}{\mathrm{2}}\:\:\Rightarrow{y}^{'} \left(\mathrm{1}\right)=\mathrm{8}\:{or}\:{y}^{'} \left(\mathrm{1}\right)=−\mathrm{5} \\ $$
Commented by math khazana by abdo last updated on 23/Jun/18
$${let}\:{f}\left({x}\right)=\frac{{sinx}}{\mathrm{1}+{x}}\:{so}\:{if}\:{x}\neq−\mathrm{1} \\ $$$${f}^{'} \left({x}\right)=\:\frac{{cosx}\left(\mathrm{1}+{x}\right)\:−{sinx}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }\:=\:\frac{{cosx}\:+{x}\:{cosx}\:−{sinx}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} } \\ $$