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

Answer

To differentiate the function \( y = \frac{x - 1}{x} \) from first principles (definition of derivative), we will use the limit definition of the derivative. The derivative of a function \( y \) at a point \( x \) is defined as:

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

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

### Step 2: Calculate \( f(x + h) \)
Now, we need to find \( f(x + h) \):

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

### Step 3: Set up the limit expression
Now we can substitute \( f(x + h) \) and \( f(x) \) into the limit definition:

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

### Step 4: Simplify the expression
To simplify the expression, we will find a common denominator for the fractions in the numerator:

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

Expanding the numerator:

\[ = \frac{(x^2 - x + hx) - (x^2 - x + hx - h)}{x(x + h)} = \frac{hx - h}{x(x + h)} = \frac{h(x - 1)}{x(x + h)} \]

### Step 5: Substitute back into the limit
Now substituting this back into our limit expression:

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

### Step 6: Evaluate the limit
As \( h \to 0 \):

\[ \frac{dy}{dx} = \frac{x - 1}{x^2} \]

### Final Result
Thus, the derivative of \( y = \frac{x - 1}{x} \) is:

\[ \frac{dy}{dx} = \frac{x - 1}{x^2} \]