Find, from first principle, the derivative of the following w.r.t. x :
`log(cosx)`

Answer

To find the derivative of the function \( f(x) = \log(\cos x) \) using the first principle of derivatives, we will follow these steps:

### Step 1: Write the definition of the derivative using the first principle
The derivative of a function \( f(x) \) from first principles is defined as:
\[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \]

### Step 2: Substitute the function into the formula
For our function \( f(x) = \log(\cos x) \), we substitute into the derivative formula:
\[ f'(x) = \lim_{h \to 0} \frac{\log(\cos(x + h)) - \log(\cos x)}{h} \]

### Step 3: Use the properties of logarithms
Using the property of logarithms that states \( \log a - \log b = \log\left(\frac{a}{b}\right) \), we can rewrite the expression:
\[ f'(x) = \lim_{h \to 0} \frac{1}{h} \log\left(\frac{\cos(x + h)}{\cos x}\right) \]

### Step 4: Simplify the cosine term
Using the cosine addition formula, we have:
\[ \cos(x + h) = \cos x \cos h - \sin x \sin h \]
Thus, we can write:
\[ \frac{\cos(x + h)}{\cos x} = \frac{\cos x \cos h - \sin x \sin h}{\cos x} = \cos h - \tan x \sin h \]
So, we have:
\[ f'(x) = \lim_{h \to 0} \frac{1}{h} \log\left(\cos h - \tan x \sin h\right) \]

### Step 5: Factor out the logarithm
Now, we can express the limit as:
\[ f'(x) = \lim_{h \to 0} \frac{1}{h} \left( \log(\cos h) + \log\left(1 - \frac{\tan x \sin h}{\cos h}\right) \right) \]

### Step 6: Evaluate the limit
As \( h \to 0 \), we know:
- \( \log(\cos h) \) approaches \( \log(1) = 0 \)
- The term \( \frac{\tan x \sin h}{\cos h} \) approaches \( 0 \) as well.

Using the property \( \lim_{h \to 0} \frac{\log(1 + u)}{h} = u \) where \( u \to 0 \):
\[ f'(x) = \lim_{h \to 0} \left( \frac{\log(\cos h)}{h} + \frac{\log(1 - \frac{\tan x \sin h}{\cos h})}{h} \right) \]

The first term approaches \( -\tan x \) and the second term approaches \( -\tan x \) as well, leading to:
\[ f'(x) = -\tan x \]

### Final Result
Thus, the derivative of \( f(x) = \log(\cos x) \) is:
\[ f'(x) = -\tan x \]