Differentiate each of the following functions from the first principal :

We have to find the derivative of $\log \cos x$ with the first principle method, so,

$f(x)=\log \cos x$

by using the first principle formula, we get,

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\log \cos (x+h)-\log \cos x}{h}$

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\log \left(\frac{\cos (x+h)}{\cos x}\right)}{h}$

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\log \left(1+\frac{\cos (x+h)}{\cos x}-1\right)}{h}$

[Adding and subtracting 1 ]

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\log \left(1+\frac{\cos (x+h)-\cos x}{\cos x}\right)}{h}$

[Rationalising]

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\log \left(1+\frac{\cos (x+h)-\cos x}{\cos x}\right)}{h} \times \frac{\frac{\cos (x+h)-\cos x}{\cos x}}{\frac{\frac{\cos (x+h)-\cos x}{\cos x}}{h}}$

[By using $\lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1$ ]

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\frac{\cos (x+h)-\cos x}{\cos x}}{h}$

$\left[\cos C-\cos D=-2 \sin \frac{C-D}{2} \sin \frac{C+D}{2}\right]$

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\frac{-2 \sin \frac{2 x+h}{2} \sin \frac{h}{2}}{\cos x}}{\frac{2 h}{2}}$ [By using $\left.\lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{-2 \sin \frac{2 x+h}{2}}{2 \cos x}$

$f^{\prime}(x)=-\tan x$