Question

Differentiate the following functions from first principles:

log cosec x

Answer 1

$$\begin{gathered} \mathrm{Let}f\left(x\right)=\log \mathrm{cosec}x \\ \Rightarrow f\left(x+h\right)=\log \mathrm{cosec}\left(x+h\right) \\ \therefore \frac{d}{dx}\left\{f\left(x\right)\right\}=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{\log \mathrm{cosec}\left(x+h\right)-\log \mathrm{cosec}x}{h} \\ =\underset{h\to 0}{\lim }\frac{\log \left\{{\displaystyle \frac{\mathrm{cosec}\left(x+h\right)}{\mathrm{cosec}x}}\right\}}{h} \\ =\underset{h\to 0}{\lim }\frac{\log \left\{1+\left({\displaystyle \frac{\sin x}{\sin \left(x+h\right)}}-1\right)\right\}}{h} \\ =\underset{h\to 0}{\lim }\left\{\frac{\log \left\{1+\left({\displaystyle \frac{\sin x-\sin \left(x+h\right)}{\sin \left(x+h\right)}}\right)\right\}}{\left\{{\displaystyle \frac{\sin x-\sin \left(x+h\right)}{\sin \left(x+h\right)}}\right\}}\right\}\frac{\left\{{\displaystyle \frac{\sin x-\sin \left(x+h\right)}{\sin \left(x+h\right)}}\right\}}{h} \\ =\underset{h\to 0}{\lim }\frac{2\cos \left({\displaystyle \frac{x+x+h}{2}}\right)\sin \left({\displaystyle \frac{x-x-h}{2}}\right)}{\sin \left(x+h\right)h}\left[\because \underset{x\to 0}{\lim }\frac{\log \left(1+x\right)}{x}=1\mathrm{and}\sin A-\sin B=2\cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)\right] \\ =\underset{h\to 0}{\lim }\frac{2\cos \left({\displaystyle \frac{2x+h}{2}}\right)}{\sin \left(x+h\right)\left(-2\right)}\left\{\frac{\sin \left(-{\displaystyle \frac{h}{2}}\right)}{-{\displaystyle \frac{h}{2}}}\right\}\left[\because \underset{x\to 0}{\lim }\frac{\sin x}{x}=1\right] \\ =-\cot x \\ \therefore \frac{d}{dx}\left(\log \mathrm{cosec}x\right)=-\cot x \end{gathered}$$