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

Answer

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

### Step 1: Define the first principle of derivatives
The first principle of derivatives states that the derivative of a function \( f(x) \) is given by:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

### Step 2: Write the function in terms of \( f(x) \)
For our function, we have:
\[ f(x) = \log(\sin x) \]

### Step 3: Substitute into the first principle formula
Now, we substitute \( f(x) \) into the formula:
\[ f'(x) = \lim_{h \to 0} \frac{\log(\sin(x+h)) - \log(\sin x)}{h} \]

### Step 4: Use the logarithmic property
Using the property of logarithms \( \log a - \log b = \log\left(\frac{a}{b}\right) \), we can rewrite the expression:
\[ f'(x) = \lim_{h \to 0} \frac{1}{h} \log\left(\frac{\sin(x+h)}{\sin x}\right) \]

### Step 5: Apply the sine addition formula
Using the sine addition formula \( \sin(x+h) = \sin x \cos h + \cos x \sin h \), we substitute:
\[ f'(x) = \lim_{h \to 0} \frac{1}{h} \log\left(\frac{\sin x \cos h + \cos x \sin h}{\sin x}\right) \]
This simplifies to:
\[ f'(x) = \lim_{h \to 0} \frac{1}{h} \log\left(\cos h + \cot x \sin h\right) \]

### Step 6: Split the logarithm
Now we can separate the logarithm:
\[ f'(x) = \lim_{h \to 0} \frac{1}{h} \left(\log(\cos h + \cot x \sin h)\right) \]

### Step 7: Evaluate the limit
As \( h \to 0 \), we know:
- \( \cos h \to 1 \)
- \( \sin h \to h \)

Thus, we can write:
\[ \cos h + \cot x \sin h \to 1 + \cot x \cdot h \]
So, we have:
\[ f'(x) = \lim_{h \to 0} \frac{1}{h} \log(1 + \cot x \cdot h) \]

Using the limit property \( \lim_{h \to 0} \frac{\log(1 + ah)}{h} = a \) for small \( h \):
\[ f'(x) = \cot x \]

### Conclusion
Thus, the derivative of \( \log(\sin x) \) is:
\[ f'(x) = \cot x \]