Differentiate, `f(x) = ax^(2) + (b)/(x)` with respect to 'x' using first principle.

Answer

To differentiate the function \( f(x) = ax^2 + \frac{b}{x} \) with respect to \( x \) using the first principle of derivatives, we will follow these steps:

### Step 1: Write the definition of the derivative using the first principle
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 \( f(x) \) into the definition
We have:
\[ f(x) = ax^2 + \frac{b}{x} \]
Now, we need to find \( f(x+h) \):
\[ f(x+h) = a(x+h)^2 + \frac{b}{x+h} \]

### Step 3: Expand \( f(x+h) \)
Expanding \( (x+h)^2 \):
\[ f(x+h) = a(x^2 + 2xh + h^2) + \frac{b}{x+h} \]
This simplifies to:
\[ f(x+h) = ax^2 + 2axh + ah^2 + \frac{b}{x+h} \]

### Step 4: Substitute \( f(x+h) \) and \( f(x) \) into the limit
Now we substitute \( f(x+h) \) and \( f(x) \) into the derivative formula:
\[ f'(x) = \lim_{h \to 0} \frac{(ax^2 + 2axh + ah^2 + \frac{b}{x+h}) - (ax^2 + \frac{b}{x})}{h} \]

### Step 5: Simplify the expression
The \( ax^2 \) terms cancel out:
\[ f'(x) = \lim_{h \to 0} \frac{2axh + ah^2 + \frac{b}{x+h} - \frac{b}{x}}{h} \]
This can be rewritten as:
\[ f'(x) = \lim_{h \to 0} \frac{2axh + ah^2 + \frac{b(x - (x+h))}{x(x+h)}}{h} \]
The term \( \frac{b}{x+h} - \frac{b}{x} \) can be simplified to:
\[ \frac{-bh}{x(x+h)} \]

### Step 6: Combine and factor out \( h \)
Now we have:
\[ f'(x) = \lim_{h \to 0} \frac{h(2ax + ah - \frac{b}{x(x+h)})}{h} \]
Cancelling \( h \):
\[ f'(x) = \lim_{h \to 0} \left( 2ax + ah - \frac{b}{x(x+h)} \right) \]

### Step 7: Take the limit as \( h \to 0 \)
As \( h \to 0 \):
\[ f'(x) = 2ax + 0 - \frac{b}{x^2} = 2ax - \frac{b}{x^2} \]

### Final Answer
Thus, the derivative of \( f(x) \) is:
\[ f'(x) = 2ax - \frac{b}{x^2} \]