Question

Differentiate the following functions from first principles:

log cos x

Answer 1

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