Derivative of log `(x+sqrt(x^(2)-1))`w.r.t.x is

Options

A. `(1)/(sqrt(x^(2-1)))`
B. `(-1)/(sqrt(x^(2)-1))`
C. `(1)/(sqrt(x^(2+1)))`
D. `(-1)/(sqrt(x^(2-1)))`

Answer

To find the derivative of the function \( y = \log(x + \sqrt{x^2 - 1}) \) with respect to \( x \), we will follow these steps:

### Step 1: Differentiate the logarithmic function
We start with the function:
\[ y = \log(x + \sqrt{x^2 - 1}) \]
Using the chain rule, the derivative of \( \log(u) \) is given by:
\[ \frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx} \]
where \( u = x + \sqrt{x^2 - 1} \).

### Step 2: Find \( \frac{du}{dx} \)
Now, we need to differentiate \( u \):
\[ u = x + \sqrt{x^2 - 1} \]
The derivative of \( u \) is:
\[ \frac{du}{dx} = \frac{d}{dx}(x) + \frac{d}{dx}(\sqrt{x^2 - 1}) \]
The derivative of \( x \) is \( 1 \). For \( \sqrt{x^2 - 1} \), we apply the chain rule:
\[ \frac{d}{dx}(\sqrt{x^2 - 1}) = \frac{1}{2\sqrt{x^2 - 1}} \cdot \frac{d}{dx}(x^2 - 1) = \frac{1}{2\sqrt{x^2 - 1}} \cdot (2x) = \frac{x}{\sqrt{x^2 - 1}} \]
Thus, we have:
\[ \frac{du}{dx} = 1 + \frac{x}{\sqrt{x^2 - 1}} \]

### Step 3: Substitute back into the derivative formula
Now substituting \( u \) and \( \frac{du}{dx} \) back into the derivative formula:
\[ \frac{dy}{dx} = \frac{1}{x + \sqrt{x^2 - 1}} \cdot \left( 1 + \frac{x}{\sqrt{x^2 - 1}} \right) \]

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

### Step 5: Final simplification
Notice that \( \sqrt{x^2 - 1} + x \) can be rewritten as:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{x^2 - 1}} \]

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

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