Question

Differentiate the following functions from first principles log cos x ?

Answer 1

\[\text{ Let } f\left( x \right) = \log \cos x\]
\[ \Rightarrow f\left( x + h \right) = \log \cos\left( x + h \right)\]
\[ \therefore \frac{d}{dx}\left\{ f\left( x \right) \right\} = \lim_{h \to 0} \frac{f\left( x + h \right) = f\left( x \right)}{h}\]
\[ = \lim_{h \to 0} \frac{\log \cos\left( x + h \right) - \log \cos x}{h}\]
\[ = \lim_{h \to 0} \frac{\log\frac{\cos\left( x + h \right)}{\cos x}}{h} \left[ \because \log A - \log B = \log\left( \frac{A}{B} \right) \right]\]
\[ = \lim_{h \to 0} \frac{\log\left[ 1 + \left\{ \frac{\cos\left( x + h \right)}{\cos x} - 1 \right\} \right]}{h}\]
\[ = \lim_{h \to 0} \frac{\log\left\{ 1 + \frac{\cos\left( x + h \right) - \cos x}{\cos x} \right\}}{\left\{ \frac{\cos\left( x + h \right) - \cos x}{\cos x} \right\}} \times \lim_{h \to 0} \left\{ \frac{\cos\left( x + h \right) - \cos x}{\cos x} \right\}\]
\[ = 1 \times \lim_{h \to 0} \frac{\cos\left( x + h \right) - \cos x}{\cos x \times h} \left[ \because \lim_{x \to 0} \frac{\log\left( 1 + x \right)}{x} = 1 \right]\]
\[ = \lim_{h \to 0} \frac{- 2\sin\left( \frac{x + h + x}{2} \right)\sin\left( \frac{x + h - x}{2} \right)}{\cos x \times h}\]
\[ = - 2 \lim_{h \to 0} \frac{\sin\left( \frac{2x + h}{2} \right) \times \sin\left( \frac{h}{2} \right)}{2\cos x \times \left( \frac{h}{2} \right)}\]
\[ = \frac{- 2\sin x}{2\cos x} \left[ \because \lim_{x \to 0} \frac{\sin x}{x} = 1 \right]\]
\[ = - \tan x\]
\[So, \frac{d}{dx}\left( \log \cos x \right) = - \tan x\]