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

Answer

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

### Step 1: Rewrite the function
First, we express \( \sqrt{x} \) in terms of exponents:
\[ f(x) = \log\left(x^{1/2} + x^{-1/2}\right) \]

### Step 2: Differentiate using the chain rule
Using the chain rule, the derivative of \( \log(u) \) is \( \frac{1}{u} \cdot \frac{du}{dx} \). Here, \( u = \sqrt{x} + \frac{1}{\sqrt{x}} \):
\[ f'(x) = \frac{1}{\sqrt{x} + \frac{1}{\sqrt{x}}} \cdot \frac{d}{dx}\left(\sqrt{x} + \frac{1}{\sqrt{x}}\right) \]

### Step 3: Differentiate \( u \)
Now, we need to differentiate \( u \):
\[ \frac{d}{dx}\left(\sqrt{x}\right) = \frac{1}{2\sqrt{x}}, \quad \text{and} \quad \frac{d}{dx}\left(\frac{1}{\sqrt{x}}\right) = -\frac{1}{2}x^{-3/2} = -\frac{1}{2\sqrt{x^3}} \]
Thus,
\[ \frac{du}{dx} = \frac{1}{2\sqrt{x}} - \frac{1}{2\sqrt{x^3}} = \frac{1}{2\sqrt{x}} - \frac{1}{2x\sqrt{x}} = \frac{1}{2\sqrt{x}} \left(1 - \frac{1}{x}\right) \]

### Step 4: Substitute back into the derivative
Now, substituting \( \frac{du}{dx} \) back into the derivative:
\[ f'(x) = \frac{1}{\sqrt{x} + \frac{1}{\sqrt{x}}} \cdot \frac{1}{2\sqrt{x}} \left(1 - \frac{1}{x}\right) \]

### Step 5: Simplify the expression
We can simplify \( f'(x) \):
\[ f'(x) = \frac{1}{\sqrt{x} + \frac{1}{\sqrt{x}}} \cdot \frac{1}{2\sqrt{x}} \cdot \frac{x - 1}{x} \]
The term \( \sqrt{x} + \frac{1}{\sqrt{x}} \) can be rewritten as \( \frac{x + 1}{\sqrt{x}} \):
\[ f'(x) = \frac{1}{\frac{x + 1}{\sqrt{x}}} \cdot \frac{1}{2\sqrt{x}} \cdot \frac{x - 1}{x} \]
This simplifies to:
\[ f'(x) = \frac{\sqrt{x}}{x + 1} \cdot \frac{1}{2\sqrt{x}} \cdot \frac{x - 1}{x} = \frac{x - 1}{2x(x + 1)} \]

### Final Result
Thus, the derivative of the function is:
\[ f'(x) = \frac{x - 1}{2x(x + 1)} \]

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