Differentiate the following w.r.t.x. `log(x+sqrt(a^(2)+x^(2)))`

Answer

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

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

### Step 2: Differentiate using the chain rule
Using the chain rule, the derivative of \( y \) with respect to \( x \) is given by:
\[ \frac{dy}{dx} = \frac{1}{x + \sqrt{a^2 + x^2}} \cdot \frac{d}{dx}(x + \sqrt{a^2 + x^2}) \]

### Step 3: Differentiate the inner function
Now we need to differentiate the inner function \( x + \sqrt{a^2 + x^2} \):
1. The derivative of \( x \) is \( 1 \).
2. 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 substitute this back into the derivative:
\[ \frac{dy}{dx} = \frac{1}{x + \sqrt{a^2 + x^2}} \left( 1 + \frac{x}{\sqrt{a^2 + x^2}} \right) \]

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

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

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

### 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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