The function log(cos x) denotes the logarithm of the cosine function. Here we will find the derivative of log(cos x) using the first principle of derivatives.

The derivative of a function $f(x)$ by the first principle of derivatives is defined to be the following limit:

$f'(x)=\dfrac{d}{dx}(f(x))$ $=\lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$ $\quad \cdots (i)$

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.

**Also Read:**

**Derivative of log(sin x) from first principle**

**Derivative of root sin x from first principle **

**Derivative of root cos x from first principle**

## FAQs

**Q1: What is the derivative of log(cosx)?**

Answer: The derivative of log(cosx) is -tanx.