Question Number 103149 by bemath last updated on 13/Jul/20
$${using}\:{first}\:{principal}\: \\ $$$${y}\:=\:\mathrm{ln}\:\left(\mathrm{sin}\:\sqrt{{x}}\right)\:\rightarrow{y}'\:=\:? \\ $$
Answered by bobhans last updated on 13/Jul/20
$${y}'\:=\:\underset{\bigtriangleup{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{ln}\left(\mathrm{sin}\:\sqrt{\mathrm{x}+\bigtriangleup{x}}\right)−\mathrm{ln}\left(\mathrm{sin}\:\sqrt{\mathrm{x}}\right)}{\bigtriangleup{x}} \\ $$$${y}'=\underset{\bigtriangleup{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{ln}\left(\frac{\mathrm{sin}\:\sqrt{\mathrm{x}+\bigtriangleup{x}}}{\mathrm{sin}\:\sqrt{{x}}}\right)}{\bigtriangleup{x}}\: \\ $$
Answered by mathmax by abdo last updated on 14/Jul/20
$$\mathrm{y}\left(\mathrm{x}\right)\:=\mathrm{ln}\left(\mathrm{sin}\left(\sqrt{\mathrm{x}}\right)\right)\:\Rightarrow\mathrm{y}^{'} \left(\mathrm{x}\right)\:=\frac{\left(\mathrm{sin}\left(\sqrt{\mathrm{x}}\right)\right)^{'} }{\mathrm{sin}\left(\sqrt{\mathrm{x}}\right)}\:=\frac{\mathrm{cos}\left(\sqrt{\mathrm{x}}\right)}{\mathrm{2}\sqrt{\mathrm{x}}\mathrm{sin}\left(\sqrt{\mathrm{x}}\right)}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{x}}}\mathrm{cotan}\left(\sqrt{\mathrm{x}}\right) \\ $$