Differentiate-ln-cosx-from-the-first-principle-

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

Differentiate, \ln (cosx) from the first principle .

Answered by ajfour last updated on 02/Apr/17

\frac{d}{d x} \ln (\cos x) = \lim_{h \to 0} \frac{\ln \cos (x + h) - \ln \cos x}{h}

= \lim_{h \to 0} \frac{\ln [\frac{\cos (x + h)}{\cos x}]}{h}

= \lim_{h \to 0} \frac{\ln [\frac{\cos x \cos h - \sin x \sin h}{\cos x}]}{h}

= \lim_{h \to 0} \frac{\ln [\cos h - \tan x \sin h]}{h}

= \lim_{h \to 0} {\frac{\ln [1 - \tan x \sin h]}{- \tan x \sin h} . \frac{(- \tan x \sin h)}{h}}

= \lim_{t \to 0} \frac{\ln (1 + t)}{t} . \lim_{h \to 0} (- \tan x) . \lim_{h \to 0} (\frac{\sin h}{h})

= (1) (- \tan x) (1) = - \tan x .

Commented by tawa last updated on 02/Apr/17

God bless you sir .

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