Differentiate `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: Differentiate the function
We start by differentiating both sides with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx} \left( \log(x + \sqrt{x^2 + a}) \right) \]

### Step 2: Apply the chain rule
Using the chain rule for the logarithmic function, 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 differentiate the inner function \( x + \sqrt{x^2 + a} \):
\[ \frac{d}{dx} (x + \sqrt{x^2 + a}) = \frac{d}{dx} x + \frac{d}{dx} \sqrt{x^2 + a} \]
The derivative of \( x \) is \( 1 \) and for \( \sqrt{x^2 + a} \), we use the chain rule:
\[ \frac{d}{dx} \sqrt{x^2 + a} = \frac{1}{2\sqrt{x^2 + a}} \cdot \frac{d}{dx} (x^2 + a) = \frac{1}{2\sqrt{x^2 + a}} \cdot 2x = \frac{x}{\sqrt{x^2 + a}} \]

### Step 4: Combine the derivatives
Now substituting back, we have:
\[ \frac{d}{dx} (x + \sqrt{x^2 + a}) = 1 + \frac{x}{\sqrt{x^2 + a}} \]

### Step 5: Substitute back into the derivative
Now substituting this result back into our derivative expression:
\[ \frac{dy}{dx} = \frac{1}{x + \sqrt{x^2 + a}} \left( 1 + \frac{x}{\sqrt{x^2 + a}} \right) \]

### Step 6: Simplify the expression
We can simplify this expression:
\[ \frac{dy}{dx} = \frac{1 + \frac{x}{\sqrt{x^2 + a}}}{x + \sqrt{x^2 + a}} = \frac{\sqrt{x^2 + a} + x}{(x + \sqrt{x^2 + a}) \sqrt{x^2 + a}} \]

### Step 7: Final simplification
Notice that \( \sqrt{x^2 + a} + x = x + \sqrt{x^2 + a} \), so we can simplify further:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{x^2 + a}} \]

### Final Answer
Thus, the derivative of the function \( y = \log(x + \sqrt{x^2 + a}) \) is:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{x^2 + a}} \]

---