`log(sqrt(x)+1/sqrt(x))`

Answer

To differentiate the function \( y = \log\left(\sqrt{x} + \frac{1}{\sqrt{x}}\right) \), we will follow these steps:

### Step 1: Rewrite the function
Let \( y = \log\left(\sqrt{x} + \frac{1}{\sqrt{x}}\right) \).

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

### Step 3: Find \( \frac{du}{dx} \)
Now we need to differentiate \( u \):
\[ u = \sqrt{x} + \frac{1}{\sqrt{x}} = x^{1/2} + x^{-1/2} \]
Differentiating \( u \):
\[ \frac{du}{dx} = \frac{1}{2}x^{-1/2} - \frac{1}{2}x^{-3/2} \]
This can be simplified to:
\[ \frac{du}{dx} = \frac{1}{2\sqrt{x}} - \frac{1}{2x\sqrt{x}} = \frac{1}{2\sqrt{x}} - \frac{1}{2x\sqrt{x}} = \frac{x - 1}{2x\sqrt{x}} \]

### Step 4: Substitute back into the derivative formula
Now substitute \( u \) and \( \frac{du}{dx} \) back into the derivative formula:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{x} + \frac{1}{\sqrt{x}}} \cdot \frac{x - 1}{2x\sqrt{x}} \]

### Step 5: Simplify the expression
The expression can be simplified:
\[ \frac{dy}{dx} = \frac{x - 1}{2x\sqrt{x}\left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)} \]
Now simplify \( \sqrt{x} + \frac{1}{\sqrt{x}} \):
\[ \sqrt{x} + \frac{1}{\sqrt{x}} = \frac{x + 1}{\sqrt{x}} \]
Thus, substituting this back gives:
\[ \frac{dy}{dx} = \frac{x - 1}{2x\sqrt{x} \cdot \frac{x + 1}{\sqrt{x}}} = \frac{x - 1}{2x(x + 1)} \]

### Final Result
So, the derivative of the function \( y = \log\left(\sqrt{x} + \frac{1}{\sqrt{x}}\right) \) is:
\[ \frac{dy}{dx} = \frac{x - 1}{2x(x + 1)} \]

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