Question

Differentiate the following functions from first principles log cosec x ?

Answer 1

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