derivative of `log (x+1/x)`

Answer

To find the derivative of the function \( f(x) = \log\left(x + \frac{1}{x}\right) \), we will apply the chain rule and the properties of logarithmic differentiation. Here’s a step-by-step solution:

### Step 1: Identify the function
We start with the function:
\[ f(x) = \log\left(x + \frac{1}{x}\right) \]

### Step 2: Differentiate using the chain rule
Using the chain rule, the derivative of \( f(x) \) is given by:
\[ f'(x) = \frac{1}{g(x)} \cdot g'(x) \]
where \( g(x) = x + \frac{1}{x} \).

### Step 3: Find \( g'(x) \)
Now, we need to find the derivative of \( g(x) \):
\[ g(x) = x + \frac{1}{x} \]
To differentiate \( g(x) \):
\[ g'(x) = 1 - \frac{1}{x^2} \]

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

### Step 5: Simplify the expression
Now we can simplify the expression:
\[ f'(x) = \frac{1 - \frac{1}{x^2}}{x + \frac{1}{x}} = \frac{1 - \frac{1}{x^2}}{\frac{x^2 + 1}{x}} = \frac{x(1 - \frac{1}{x^2})}{x^2 + 1} \]
This simplifies to:
\[ f'(x) = \frac{x - \frac{1}{x}}{x^2 + 1} \]

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

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