Differentiate the following from ab-initio (or from definition) :
`(x-2)/(x+3), x ne -3`

Answer

To differentiate the function \( f(x) = \frac{x - 2}{x + 3} \) from first principles, we will use 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 \( f(x + h) \):

\[ f(x + h) = \frac{(x + h) - 2}{(x + h) + 3} = \frac{x + h - 2}{x + h + 3} \]

### Step 2: Set up the difference quotient

Now we set up the difference quotient:

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

### Step 3: Combine the fractions in the numerator

To combine the fractions in the numerator, we find a common denominator:

\[ \frac{x + h - 2}{x + h + 3} - \frac{x - 2}{x + 3} = \frac{(x + h - 2)(x + 3) - (x - 2)(x + h + 3)}{(x + h + 3)(x + 3)} \]

### Step 4: Expand the numerator

Now we expand the numerator:

\[ = (x + h - 2)(x + 3) - (x - 2)(x + h + 3) \]

Expanding both terms:

1. \( (x + h - 2)(x + 3) = x^2 + 3x + hx + 3h - 2x - 6 = x^2 + (3 - 2)x + hx + 3h - 6 \)
2. \( (x - 2)(x + h + 3) = x^2 + hx + 3x - 2x - 6 = x^2 + (3 - 2)x + hx - 6 \)

Combining these gives:

\[ = (x^2 + (3 - 2)x + hx + 3h - 6) - (x^2 + (3 - 2)x + hx - 6) \]

### Step 5: Simplify the numerator

The \( x^2 \) and \( -6 \) terms cancel out, leaving:

\[ = 3h \]

### Step 6: Substitute back into the limit

Now substituting back into the limit gives:

\[ f'(x) = \lim_{h \to 0} \frac{3h}{h \cdot (x + h + 3)(x + 3)} \]

### Step 7: Cancel \( h \)

Cancelling \( h \) from the numerator and denominator:

\[ = \lim_{h \to 0} \frac{3}{(x + h + 3)(x + 3)} \]

### Step 8: Evaluate the limit

Now we evaluate the limit as \( h \to 0 \):

\[ = \frac{3}{(x + 0 + 3)(x + 3)} = \frac{3}{(x + 3)^2} \]

### Final Result

Thus, the derivative of \( f(x) = \frac{x - 2}{x + 3} \) is:

\[ f'(x) = \frac{3}{(x + 3)^2} \]

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