$f'(x)=\dfrac{d}{dx}(f(x))$ $=\lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$ $\quad \cdots (i)$
Here the symbol $’$ denotes the derivative of a function.
Using this first principle, we will now find the derivative of $\log(\cos x)$.
Derivative of log(cos x) by First Principle
In the above rule (i) of the first principle of
the derivative, we will take $f(x)=\log(\cos x)$. So the derivative of
$\log(\cos x)$ by the first principle is
$\dfrac{d}{dx}(\log(\cos x)) = (\log \cos
x)’$$=\lim\limits_{h \to 0} \dfrac{\log(\cos(x+h))-\log(\cos x)}{h}$
To find this limit, we will proceed as follows:
Step 1: At first, we will apply the formula of
$\log a -\log b =\log (a/b)$. So the above limit is
$=\lim\limits_{h \to 0} \dfrac{\log(\dfrac{\cos(x+h)}{\cos
x})}{h}$
Step 2: Next, using the trigonometric formula of
$\cos(a+b)=\cos a \cos b-\sin a \sin b$, we get
$=\lim\limits_{h \to 0}$ $\dfrac{\log(\dfrac{\cos x \cos h
-\sin x \sin h}{\cos x})}{h}$
$=\lim\limits_{h \to 0}$ $\dfrac{\log(\dfrac{\cos x \cos
h}{\cos x} -\dfrac{\sin x \sin h}{\cos x})}{h}$
$=\lim\limits_{h \to 0}$ $\dfrac{\log(\cos h – \tan x \sin
h)}{h} \quad$ as we know that $\dfrac{\sin x}{\cos x} =\tan x$.
Step 3: As $\cos h$ tends to $1$ when $h \to 0$,
from the above step we obtain that the derivative of $\log \sin x$ is
$=\lim\limits_{h \to 0}$ $\dfrac{\log(1 – \tan x \sin
h)}{h}$
$=\lim\limits_{h \to 0}$ $[\dfrac{\log(1 – \tan x \sin
h)}{-\tan x \sin h}$ $\times \dfrac{-\tan x \sin h}{h}]$
Let $z=-\tan x \sin h$. Then $z \to 0$ when $h \to 0$.
$=\lim\limits_{z \to 0}$ $\dfrac{\log(1 + z)}{z}$ $\times
\lim\limits_{h \to 0} \dfrac{-\tan x \sin h}{h}$
$=1 \times (-\tan x) \lim\limits_{h \to 0} \dfrac{\sin
h}{h}$ as the limit of $\log(1+z)/z$ is one when z tends to zero.
$=-\tan x \times 1$ as $\lim\limits_{h \to 0} \dfrac{\sin
h}{h} =1$
$=-\tan x$
Thus the derivative of log(cos x) is -tan x, and this is obtained by the first principle of derivatives.