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Question Number 11843 by tawa last updated on 02/Apr/17

$$\mathrm{Differentiate},\ln \left(\mathrm{cosx}\right)\mathrm{from}\mathrm{the}\mathrm{first}\mathrm{principle}.$$

Answered by ajfour last updated on 02/Apr/17

$$\frac{d}{dx}\ln \left(\cos x\right)=\underset{h\to 0}{\lim }\frac{\ln \cos (x+h)-\ln \cos x}{h}$$$$=\underset{h\to 0}{\lim }\frac{\ln \left[\frac{\cos (x+h)}{\cos x}\right]}{h}$$$$=\underset{h\to 0}{\lim }\frac{\ln \left[\frac{\cos x\cos h-\sin x\sin h}{\cos x}\right]}{h}$$$$=\underset{h\to 0}{\lim }\frac{\ln [\cos h-\tan x\sin h]}{h}$$$$=\underset{h\to 0}{\lim }\{\frac{\ln [1-\tan x\sin h]}{-\tan x\sin h}.\frac{(-\tan x\sin h)}{h}\}$$$$=\underset{t\to 0}{\lim }\frac{\ln (1+t)}{t}.\underset{h\to 0}{\lim }(-\tan x).\underset{h\to 0}{\lim }\left(\frac{\sin h}{h}\right)$$$$=\left(1\right)(-\tan x)\left(1\right)=-\tan x.$$

Commented by tawa last updated on 02/Apr/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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