`log(x+sqrt(x^(2)+a))`

Explanation

To differentiate the function \( y = \log(x + \sqrt{x^2 + a}) \), we will follow these steps:

### Step 1: Identify the function
Let
\[ y = \log(x + \sqrt{x^2 + a}) \]

### Step 2: Differentiate using the chain rule
We will differentiate \( y \) with respect to \( x \). Using the chain rule, we have:
\[ \frac{dy}{dx} = \frac{1}{x + \sqrt{x^2 + a}} \cdot \frac{d}{dx}(x + \sqrt{x^2 + a}) \]

### Step 3: Differentiate the inner function
Now, we need to differentiate the inner function \( x + \sqrt{x^2 + a} \):
\[ \frac{d}{dx}(x + \sqrt{x^2 + a}) = 1 + \frac{1}{2\sqrt{x^2 + a}} \cdot \frac{d}{dx}(x^2 + a) \]
Since \( \frac{d}{dx}(x^2 + a) = 2x \), we have:
\[ \frac{d}{dx}(x + \sqrt{x^2 + a}) = 1 + \frac{1}{2\sqrt{x^2 + a}} \cdot 2x = 1 + \frac{x}{\sqrt{x^2 + a}} \]

### Step 4: Substitute back into the derivative
Now substituting this back into our expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{1}{x + \sqrt{x^2 + a}} \left( 1 + \frac{x}{\sqrt{x^2 + a}} \right) \]

### Step 5: Simplify the expression
We can simplify this further:
\[ \frac{dy}{dx} = \frac{1 + \frac{x}{\sqrt{x^2 + a}}}{x + \sqrt{x^2 + a}} \]
To combine the terms in the numerator:
\[ \frac{dy}{dx} = \frac{\sqrt{x^2 + a} + x}{(x + \sqrt{x^2 + a}) \sqrt{x^2 + a}} \]

### Step 6: Final expression
Thus, the final result for the derivative is:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{x^2 + a}} \]