Find, from first principle, the derivative of the following w.r.t. x :
`cos(logx),"where "xgt0`.

Answer

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

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

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

### Step 3: Apply the cosine difference identity
Using the identity \( \cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \), we set \( A = \log(x+h) \) and \( B = \log x \):
\[ f'(x) = \lim_{h \to 0} \frac{-2 \sin\left(\frac{\log(x+h) + \log x}{2}\right) \sin\left(\frac{\log(x+h) - \log x}{2}\right)}{h} \]

### Step 4: Simplify the logarithmic expressions
Using the property of logarithms, we have:
\[ \log(x+h) = \log x + \log\left(1 + \frac{h}{x}\right) \]
Thus,
\[ \log(x+h) - \log x = \log\left(1 + \frac{h}{x}\right) \]
So,
\[ f'(x) = \lim_{h \to 0} \frac{-2 \sin\left(\frac{\log x + \log\left(1 + \frac{h}{x}\right)}{2}\right) \sin\left(\frac{\log\left(1 + \frac{h}{x}\right)}{2}\right)}{h} \]

### Step 5: Evaluate the limit
As \( h \to 0 \), \( \log\left(1 + \frac{h}{x}\right) \) approaches \( \frac{h}{x} \). Therefore, we can rewrite:
\[ \sin\left(\frac{\log\left(1 + \frac{h}{x}\right)}{2}\right) \approx \frac{\frac{h}{2x}}{2} = \frac{h}{4x} \]
Thus, we have:
\[ f'(x) = \lim_{h \to 0} \frac{-2 \sin\left(\frac{\log x + \log(1)}{2}\right) \cdot \frac{h}{4x}}{h} \]

### Step 6: Cancel \( h \) and finalize the derivative
The \( h \) terms cancel out:
\[ f'(x) = -\frac{1}{2x} \cdot 2 \sin\left(\log x\right) \]
Thus, we arrive at:
\[ f'(x) = -\frac{\sin(\log x)}{x} \]

### Final Result
The derivative of \( f(x) = \cos(\log x) \) is:
\[ f'(x) = -\frac{\sin(\log x)}{x} \]