If y= `log (x+sqrt(x^2-1))`, then dy/dx=

Answer

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

### Step 1: Differentiate the function
We start with the function:
\[ y = \log(x + \sqrt{x^2 - 1}) \]
To differentiate \( y \) with respect to \( x \), we apply the chain rule and the derivative of the logarithm function.

### Step 2: Apply the chain rule
The derivative of \( \log(u) \) is \( \frac{1}{u} \cdot \frac{du}{dx} \). Here, \( u = x + \sqrt{x^2 - 1} \). Thus, we have:
\[ \frac{dy}{dx} = \frac{1}{x + \sqrt{x^2 - 1}} \cdot \frac{d}{dx}(x + \sqrt{x^2 - 1}) \]

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

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

### Step 5: Simplify the expression
To simplify \( 1 + \frac{x}{\sqrt{x^2 - 1}} \), we can write it as:
\[ 1 + \frac{x}{\sqrt{x^2 - 1}} = \frac{\sqrt{x^2 - 1}}{\sqrt{x^2 - 1}} + \frac{x}{\sqrt{x^2 - 1}} = \frac{\sqrt{x^2 - 1} + x}{\sqrt{x^2 - 1}} \]
Thus, we have:
\[ \frac{dy}{dx} = \frac{\sqrt{x^2 - 1} + x}{(x + \sqrt{x^2 - 1}) \sqrt{x^2 - 1}} \]

### Step 6: Final simplification
Notice that \( \sqrt{x^2 - 1} + x \) and \( x + \sqrt{x^2 - 1} \) are the same. Therefore, we can cancel these terms:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{x^2 - 1}} \]

### Final Result
Thus, we conclude that:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{x^2 - 1}} \]

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