Differentiate the following from ab-initio (or from definition) :
`x+1/x (x ne 0)`

Answer

To differentiate the function \( f(x) = x + \frac{1}{x} \) from first principles, we will use the definition of the derivative. The derivative \( f'(x) \) is defined as:

\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

### Step 1: Define the function
Let \( f(x) = x + \frac{1}{x} \).

### Step 2: Calculate \( f(x+h) \)
We need to find \( f(x+h) \):

\[ f(x+h) = (x+h) + \frac{1}{x+h} = x + h + \frac{1}{x+h} \]

### Step 3: Set up the difference quotient
Now, we can set up the difference quotient:

\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{\left( x + h + \frac{1}{x+h} \right) - \left( x + \frac{1}{x} \right)}{h} \]

### Step 4: Simplify the expression
This simplifies to:

\[ f'(x) = \lim_{h \to 0} \frac{h + \frac{1}{x+h} - \frac{1}{x}}{h} \]

### Step 5: Combine the fractions
Now, we need to combine the fractions in the numerator:

\[ \frac{1}{x+h} - \frac{1}{x} = \frac{x - (x+h)}{x(x+h)} = \frac{-h}{x(x+h)} \]

Substituting this back into the limit gives:

\[ f'(x) = \lim_{h \to 0} \frac{h - \frac{h}{x(x+h)}}{h} \]

### Step 6: Factor out \( h \)
We can factor out \( h \) from the numerator:

\[ f'(x) = \lim_{h \to 0} \left( 1 - \frac{1}{x(x+h)} \right) \]

### Step 7: Evaluate the limit
Now, we can evaluate the limit as \( h \) approaches 0:

\[ f'(x) = 1 - \frac{1}{x^2} \]

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

\[ f'(x) = 1 - \frac{1}{x^2} \]