Differentiate the Following Functions from First Principles Log Cos X ? - Mathematics

Question

Differentiate the following functions from first principles log cos x ?

Solution

$Let f (x) = \log \cos x$
$\Rightarrow f (x + h) = \log \cos (x + h)$
$∴ \frac{d}{d x} {f (x)} = lim_{h \to 0} \frac{f (x + h) = f (x)}{h}$
$= lim_{h \to 0} \frac{\log \cos (x + h) - \log \cos x}{h}$
$= lim_{h \to 0} \frac{\log \frac{\cos (x + h)}{\cos x}}{h} [∵ \log A - \log B = \log (\frac{A}{B})]$
$= lim_{h \to 0} \frac{\log [1 + {\frac{\cos (x + h)}{\cos x} - 1}]}{h}$
$= lim_{h \to 0} \frac{\log {1 + \frac{\cos (x + h) - \cos x}{\cos x}}}{{\frac{\cos (x + h) - \cos x}{\cos x}}} \times lim_{h \to 0} {\frac{\cos (x + h) - \cos x}{\cos x}}$
$= 1 \times lim_{h \to 0} \frac{\cos (x + h) - \cos x}{\cos x \times h} [∵ lim_{x \to 0} \frac{\log (1 + x)}{x} = 1]$
$= lim_{h \to 0} \frac{- 2 \sin (\frac{x + h + x}{2}) \sin (\frac{x + h - x}{2})}{\cos x \times h}$
$= - 2 lim_{h \to 0} \frac{\sin (\frac{2 x + h}{2}) \times \sin (\frac{h}{2})}{2 \cos x \times (\frac{h}{2})}$
$= \frac{- 2 \sin x}{2 \cos x} [∵ lim_{x \to 0} \frac{\sin x}{x} = 1]$
$= - \tan x$
$S o, \frac{d}{d x} (\log \cos x) = - \tan x$

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Simple Problems on Applications of Derivatives

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Chapter 11: Differentiation - Exercise 11.01 [Page 17]

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Chapter 11 Differentiation
Exercise 11.01 | Q 6 | Page 17

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Differentiate the Following Functions from First Principles Log Cos X ? - Mathematics

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