Using first principles, prove that `(d)/( dx) ((1)/( f(x))) = (-f' (x))/( {f (x) }^2)`.

Answer

To prove that \(\frac{d}{dx} \left(\frac{1}{f(x)}\right) = -\frac{f'(x)}{(f(x))^2}\) using first principles, we will follow these steps:

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

### Step 2: Apply the definition to \(\frac{1}{f(x)}\)
We need to find the derivative of \(\frac{1}{f(x)}\). Thus, we can write:
\[ \frac{d}{dx} \left(\frac{1}{f(x)}\right) = \lim_{h \to 0} \frac{\frac{1}{f(x+h)} - \frac{1}{f(x)}}{h} \]

### Step 3: Simplify the expression
To simplify the expression inside the limit, we can find a common denominator:
\[ \frac{1}{f(x+h)} - \frac{1}{f(x)} = \frac{f(x) - f(x+h)}{f(x)f(x+h)} \]
Thus, we have:
\[ \frac{d}{dx} \left(\frac{1}{f(x)}\right) = \lim_{h \to 0} \frac{f(x) - f(x+h)}{h \cdot f(x)f(x+h)} \]

### Step 4: Factor out the negative sign
We can factor out the negative sign from the numerator:
\[ = \lim_{h \to 0} \frac{- (f(x+h) - f(x))}{h \cdot f(x)f(x+h)} \]

### Step 5: Apply the limit
Now, we can separate the limit:
\[ = -\lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \cdot \frac{1}{f(x)f(x+h)} \]
From the definition of the derivative, we know that:
\[ \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = f'(x) \]
So we can substitute this into our expression:
\[ = -f'(x) \cdot \lim_{h \to 0} \frac{1}{f(x)f(x+h)} \]

### Step 6: Evaluate the limit
As \(h\) approaches 0, \(f(x+h)\) approaches \(f(x)\):
\[ = -f'(x) \cdot \frac{1}{f(x)f(x)} = -\frac{f'(x)}{(f(x))^2} \]

### Conclusion
Thus, we have shown that:
\[ \frac{d}{dx} \left(\frac{1}{f(x)}\right) = -\frac{f'(x)}{(f(x))^2} \]