Question Number 11843 by tawa last updated on 02/Apr/17
$$\mathrm{Differentiate},\:\:\mathrm{ln}\left(\mathrm{cosx}\right)\:\:\:\mathrm{from}\:\mathrm{the}\:\mathrm{first}\:\mathrm{principle}. \\ $$
Answered by ajfour last updated on 02/Apr/17
![(d/dx)ln (cos x)=lim_(h→0) ((ln cos (x+h)−ln cos x)/h) =lim_(h→0) ((ln [ ((cos (x+h))/(cos x)) ])/h) =lim_(h→0) ((ln [((cos xcos h−sin xsin h)/(cos x)) ])/h) =lim_(h→0) ((ln [cos h−tan xsin h ])/h) =lim_(h→0) { ((ln [1−tan xsin h ])/(−tan xsin h)).(((−tan xsin h))/h)} =lim_(t→0) ((ln (1+t))/t).lim_(h→0) (−tan x).lim_(h→0) (((sin h)/h)) = (1)(−tan x)(1) = −tan x .](Q11845.png)
$$\frac{{d}}{{dx}}\mathrm{ln}\:\left(\mathrm{cos}\:{x}\right)=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\mathrm{cos}\:\left({x}+{h}\right)−\mathrm{ln}\:\mathrm{cos}\:{x}}{{h}} \\ $$$$\:=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left[\:\frac{\mathrm{cos}\:\left({x}+{h}\right)}{\mathrm{cos}\:{x}}\:\right]}{{h}} \\ $$$$=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left[\frac{\mathrm{cos}\:{x}\mathrm{cos}\:{h}−\mathrm{sin}\:{x}\mathrm{sin}\:{h}}{\mathrm{cos}\:{x}}\:\right]}{{h}} \\ $$$$=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left[\mathrm{cos}\:{h}−\mathrm{tan}\:{x}\mathrm{sin}\:{h}\:\right]}{{h}} \\ $$$$=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\:\frac{\mathrm{ln}\:\left[\mathrm{1}−\mathrm{tan}\:{x}\mathrm{sin}\:{h}\:\right]}{−\mathrm{tan}\:{x}\mathrm{sin}\:{h}}.\frac{\left(−\mathrm{tan}\:{x}\mathrm{sin}\:{h}\right)}{{h}}\right\} \\ $$$$=\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{ln}\:\left(\mathrm{1}+{t}\right)}{{t}}.\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(−\mathrm{tan}\:{x}\right).\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{sin}\:{h}}{{h}}\right) \\ $$$$=\:\:\left(\mathrm{1}\right)\left(−\mathrm{tan}\:{x}\right)\left(\mathrm{1}\right)\:\:=\:−\mathrm{tan}\:{x}\:. \\ $$
Commented by tawa last updated on 02/Apr/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$