Differentiate from first principles:
12. `(x^2+1)/( x)`

Answer

To differentiate the function \( f(x) = \frac{x^2 + 1}{x} \) from first principles, we will follow the definition of the derivative:

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

### Step 1: Calculate \( f(x+h) \)

First, we need to find \( f(x+h) \):

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

Expanding \( (x+h)^2 \):

\[ f(x+h) = \frac{x^2 + 2xh + h^2 + 1}{x+h} \]

### Step 2: Substitute \( f(x) \) and \( f(x+h) \) into the derivative formula

Now we substitute \( f(x+h) \) and \( f(x) \) into the derivative formula:

\[ f'(x) = \lim_{h \to 0} \frac{\frac{x^2 + 2xh + h^2 + 1}{x+h} - \frac{x^2 + 1}{x}}{h} \]

### Step 3: Simplify the expression

To simplify the expression, we need a common denominator for the two fractions in the numerator:

\[ f'(x) = \lim_{h \to 0} \frac{(x^2 + 2xh + h^2 + 1)x - (x^2 + 1)(x+h)}{h(x+h)x} \]

Expanding the numerator:

1. The first term:
\[ (x^2 + 2xh + h^2 + 1)x = x^3 + 2x^2h + xh^2 + x \]

2. The second term:
\[ (x^2 + 1)(x + h) = x^3 + x^2h + x + h \]

Now, substituting these back into the limit:

\[ f'(x) = \lim_{h \to 0} \frac{(x^3 + 2x^2h + xh^2 + x) - (x^3 + x^2h + x + h)}{h(x+h)x} \]

### Step 4: Combine like terms

Now, combining like terms in the numerator:

\[ = \lim_{h \to 0} \frac{(2x^2h - x^2h - h)}{h(x+h)x} \]
\[ = \lim_{h \to 0} \frac{(x^2h - h)}{h(x+h)x} \]
\[ = \lim_{h \to 0} \frac{h(x^2 - 1)}{h(x+h)x} \]

### Step 5: Cancel \( h \) and take the limit

Cancel \( h \) from numerator and denominator:

\[ = \lim_{h \to 0} \frac{x^2 - 1}{(x+h)x} \]

Now, as \( h \) approaches 0, we have:

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

### Final Result

Thus, the derivative of the function \( f(x) \) is:

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