If `y = log (x + sqrt(x^(2) + a^(2))) " then " (dy)/(dx) =` ?

Options

A. `(1)/(2(x + sqrt(x^(2) + a^(2))))`
B. `(-1)/(sqrt(x^(2) + a^(2)))`
C. `(1)/(sqrt(x^(2) + a^(2)))`
D. none of these

Explanation

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

### Step 1: Differentiate the function
We start with the function:
\[ y = \log(x + \sqrt{x^2 + a^2}) \]
To find \( \frac{dy}{dx} \), we apply the derivative of the logarithm:
\[ \frac{dy}{dx} = \frac{1}{x + \sqrt{x^2 + a^2}} \cdot \frac{d}{dx}(x + \sqrt{x^2 + a^2}) \]

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

### Step 3: Substitute back into the derivative
Now we substitute this back into our derivative:
\[ \frac{dy}{dx} = \frac{1}{x + \sqrt{x^2 + a^2}} \cdot \left(1 + \frac{x}{\sqrt{x^2 + a^2}}\right) \]

### Step 4: Simplify the expression
Now we can simplify the expression:
\[ \frac{dy}{dx} = \frac{1 + \frac{x}{\sqrt{x^2 + a^2}}}{x + \sqrt{x^2 + a^2}} \]
To combine the terms in the numerator:
\[ \frac{dy}{dx} = \frac{\sqrt{x^2 + a^2} + x}{\sqrt{x^2 + a^2}(x + \sqrt{x^2 + a^2})} \]

### Final Result
Thus, the derivative is:
\[ \frac{dy}{dx} = \frac{\sqrt{x^2 + a^2} + x}{\sqrt{x^2 + a^2}(x + \sqrt{x^2 + a^2})} \]