Differentiate `log(x+sqrt(a^2+x^2))` with respect to `x` :

Explanation

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

### Step 1: Write the function
We start with the function:
\[ y = \log(x + \sqrt{a^2 + x^2}) \]

### Step 2: Differentiate using the chain rule
Using the chain rule, the derivative of \( \log(u) \) is \( \frac{1}{u} \cdot \frac{du}{dx} \). Here, \( u = x + \sqrt{a^2 + x^2} \).

Thus, we have:
\[ \frac{dy}{dx} = \frac{1}{x + \sqrt{a^2 + x^2}} \cdot \frac{d}{dx}(x + \sqrt{a^2 + x^2}) \]

### Step 3: Differentiate the inner function
Now we need to differentiate \( u = x + \sqrt{a^2 + x^2} \):
- The derivative of \( x \) is \( 1 \).
- For \( \sqrt{a^2 + x^2} \), we use the chain rule:
\[ \frac{d}{dx}(\sqrt{a^2 + x^2}) = \frac{1}{2\sqrt{a^2 + x^2}} \cdot \frac{d}{dx}(a^2 + x^2) = \frac{1}{2\sqrt{a^2 + x^2}} \cdot (0 + 2x) = \frac{x}{\sqrt{a^2 + x^2}} \]

Combining these, we get:
\[ \frac{d}{dx}(x + \sqrt{a^2 + x^2}) = 1 + \frac{x}{\sqrt{a^2 + x^2}} \]

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

### Step 5: Simplify the expression
To simplify:
\[ \frac{dy}{dx} = \frac{1 + \frac{x}{\sqrt{a^2 + x^2}}}{x + \sqrt{a^2 + x^2}} \]

We can combine the terms in the numerator:
\[ = \frac{\sqrt{a^2 + x^2} + x}{\sqrt{a^2 + x^2}(x + \sqrt{a^2 + x^2})} \]

### Step 6: Final simplification
Notice that \( \sqrt{a^2 + x^2} + x = \sqrt{a^2 + x^2} + x \) is the same as \( x + \sqrt{a^2 + x^2} \). Thus, we can simplify:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{a^2 + x^2}} \]

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

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