Derivative of log(cos x) by First Principle

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.

Derivative of log(cos x) by first principle

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 $\log(\cos x)$.

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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.

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