`log (x+sqrt(x^2-1))`

Answer

To differentiate the function \( f(x) = \log(x + \sqrt{x^2 - 1}) \) with respect to \( x \), we will use the chain rule and the properties of logarithmic differentiation. Here’s a step-by-step solution:

### Step 1: Identify the function
We have:
\[ f(x) = \log(x + \sqrt{x^2 - 1}) \]

### Step 2: Differentiate using the chain rule
Using the chain rule, the derivative of \( \log(u) \) is given by:
\[ \frac{d}{dx} \log(u) = \frac{1}{u} \cdot \frac{du}{dx} \]
where \( u = x + \sqrt{x^2 - 1} \).

### Step 3: Find \( \frac{du}{dx} \)
Now, we need to differentiate \( u \):
\[ u = x + \sqrt{x^2 - 1} \]
Differentiating \( u \):
\[ \frac{du}{dx} = 1 + \frac{d}{dx}(\sqrt{x^2 - 1}) \]
Using the chain rule for the square root:
\[ \frac{d}{dx}(\sqrt{x^2 - 1}) = \frac{1}{2\sqrt{x^2 - 1}} \cdot \frac{d}{dx}(x^2 - 1) = \frac{1}{2\sqrt{x^2 - 1}} \cdot 2x = \frac{x}{\sqrt{x^2 - 1}} \]
Thus,
\[ \frac{du}{dx} = 1 + \frac{x}{\sqrt{x^2 - 1}} \]

### Step 4: Substitute back into the derivative
Now substituting \( u \) and \( \frac{du}{dx} \) back into the derivative of \( f(x) \):
\[ \frac{df}{dx} = \frac{1}{u} \cdot \frac{du}{dx} = \frac{1}{x + \sqrt{x^2 - 1}} \left( 1 + \frac{x}{\sqrt{x^2 - 1}} \right) \]

### Step 5: Simplify the expression
Now we simplify:
\[ \frac{df}{dx} = \frac{1 + \frac{x}{\sqrt{x^2 - 1}}}{x + \sqrt{x^2 - 1}} = \frac{\sqrt{x^2 - 1} + x}{(x + \sqrt{x^2 - 1}) \sqrt{x^2 - 1}} \]

### Step 6: Final result
Thus, the derivative of the function is:
\[ \frac{df}{dx} = \frac{1}{\sqrt{x^2 - 1}} \]