Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

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

$${y}+\bigtriangleup{y}={ln}\sqrt{{cos}\left({x}+\bigtriangleup{x}\right)}\:\: \\ $$$$\bigtriangleup{y}=\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\left({cos}\left({x}+\bigtriangleup{x}\right)\right\}−\frac{\mathrm{1}}{\mathrm{2}}\left\{{lncosx}\right\}\right. \\ $$$$\bigtriangleup{y}=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left\{\frac{{cos}\left({x}+\bigtriangleup{x}\right)}{{cosx}}\right\} \\ $$$$\frac{{dy}}{{dx}}=\underset{\bigtriangleup{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\bigtriangleup{y}}{\bigtriangleup{x}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{\bigtriangleup{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{2}}×\frac{{ln}\left\{\mathrm{1}+\frac{{cos}\left({x}+\bigtriangleup{x}\right)}{{cosx}}−\mathrm{1}\right\}}{\left\{\frac{{cos}\left({x}+\bigtriangleup{x}\right)}{{cosx}}−\mathrm{1}\right\}}×\frac{{cos}\left({x}+\bigtriangleup{x}\right)−{cosx}}{{cosx}×\bigtriangleup{x}} \\ $$$${let}\:{t}=\frac{{cos}\left({x}+\bigtriangleup{x}\right)}{{cosx}}−\mathrm{1} \\ $$$${when}\:\bigtriangleup{x}\rightarrow\mathrm{0}\:\:{t}\rightarrow\mathrm{0} \\ $$$${so}\: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:×\frac{{ln}\left(\mathrm{1}+{t}\right)}{{t}}×\underset{\bigtriangleup{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}{sin}\left({x}+\frac{\bigtriangleup{x}}{\mathrm{2}}\right){sin}\left(\left(−\frac{\bigtriangleup{x}}{\mathrm{2}}\right)\right.}{{cosx}×\left(\frac{−\bigtriangleup{x}}{\mathrm{2}}\right)×\frac{−\mathrm{2}}{}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{1}×\frac{\mathrm{2}{sinx}×\mathrm{1}}{−\mathrm{2}{cosx}}=\frac{\mathrm{1}}{\mathrm{2}}×−{tanx} \\ $$$${recheck} \\ $$$$\frac{{d}\left\{\frac{\mathrm{1}}{\mathrm{2}}{lncosx}\right\}}{{dx}}=\frac{\mathrm{1}}{\mathrm{2}}×−{tanx} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com