Menu Close

use-the-first-principle-y-ln-cos-x-




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

Leave a Reply

Your email address will not be published. Required fields are marked *