Question Number 43854 by peter frank last updated on 16/Sep/18
$${use}\:{the}\:{first}\:{principle}\:{y}=\mathrm{ln}\:\sqrt{\mathrm{cos}\:{x}} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 16/Sep/18
$${y}+\bigtriangleup{y}={ln}\sqrt{{cos}\left({x}+\bigtriangleup{x}\right)}\:\: \\ $$$$\bigtriangleup{y}=\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\left({cos}\left({x}+\bigtriangleup{x}\right)\right\}−\frac{\mathrm{1}}{\mathrm{2}}\left\{{lncosx}\right\}\right. \\ $$$$\bigtriangleup{y}=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left\{\frac{{cos}\left({x}+\bigtriangleup{x}\right)}{{cosx}}\right\} \\ $$$$\frac{{dy}}{{dx}}=\underset{\bigtriangleup{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\bigtriangleup{y}}{\bigtriangleup{x}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{\bigtriangleup{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{2}}×\frac{{ln}\left\{\mathrm{1}+\frac{{cos}\left({x}+\bigtriangleup{x}\right)}{{cosx}}−\mathrm{1}\right\}}{\left\{\frac{{cos}\left({x}+\bigtriangleup{x}\right)}{{cosx}}−\mathrm{1}\right\}}×\frac{{cos}\left({x}+\bigtriangleup{x}\right)−{cosx}}{{cosx}×\bigtriangleup{x}} \\ $$$${let}\:{t}=\frac{{cos}\left({x}+\bigtriangleup{x}\right)}{{cosx}}−\mathrm{1} \\ $$$${when}\:\bigtriangleup{x}\rightarrow\mathrm{0}\:\:{t}\rightarrow\mathrm{0} \\ $$$${so}\: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:×\frac{{ln}\left(\mathrm{1}+{t}\right)}{{t}}×\underset{\bigtriangleup{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}{sin}\left({x}+\frac{\bigtriangleup{x}}{\mathrm{2}}\right){sin}\left(\left(−\frac{\bigtriangleup{x}}{\mathrm{2}}\right)\right.}{{cosx}×\left(\frac{−\bigtriangleup{x}}{\mathrm{2}}\right)×\frac{−\mathrm{2}}{}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{1}×\frac{\mathrm{2}{sinx}×\mathrm{1}}{−\mathrm{2}{cosx}}=\frac{\mathrm{1}}{\mathrm{2}}×−{tanx} \\ $$$${recheck} \\ $$$$\frac{{d}\left\{\frac{\mathrm{1}}{\mathrm{2}}{lncosx}\right\}}{{dx}}=\frac{\mathrm{1}}{\mathrm{2}}×−{tanx} \\ $$