Find, from first principles, the derivative of the following w.r.t. x :
`(-x)^(-1)`

Answer

To find the derivative of the function \( f(x) = (-x)^{-1} \) using first principles, we will follow these steps:

### Step 1: Write the definition of the derivative using first principles.
The derivative of a function \( f(x) \) at a point \( x \) is defined as:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

### Step 2: Substitute the function into the definition.
For our function \( f(x) = (-x)^{-1} \):
\[ f'(x) = \lim_{h \to 0} \frac{(-x-h)^{-1} - (-x)^{-1}}{h} \]

### Step 3: Simplify the expression inside the limit.
We can rewrite \( (-x-h)^{-1} \) and \( (-x)^{-1} \):
\[ f'(x) = \lim_{h \to 0} \frac{-\frac{1}{x+h} + \frac{1}{x}}{h} \]
This simplifies to:
\[ f'(x) = \lim_{h \to 0} \frac{-\frac{1}{x+h} + \frac{1}{x}}{h} = \lim_{h \to 0} \frac{\frac{-x + (x+h)}{x(x+h)}}{h} \]
\[ = \lim_{h \to 0} \frac{h}{h \cdot x(x+h)} \]

### Step 4: Cancel \( h \) in the numerator and denominator.
\[ = \lim_{h \to 0} \frac{1}{x(x+h)} \]

### Step 5: Evaluate the limit as \( h \) approaches 0.
Substituting \( h = 0 \):
\[ f'(x) = \frac{1}{x(x+0)} = \frac{1}{x^2} \]

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