Differentiate `(ax+b)/( ax+d)` with respect to `x` from definition.

Answer

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

### Step 1: Define the function and \( f(x + h) \)
Let \( f(x) = \frac{ax + b}{ax + d} \).

Now, we need to find \( f(x + h) \):
\[ f(x + h) = \frac{a(x + h) + b}{a(x + h) + d} = \frac{ax + ah + b}{ax + ah + d} \]

### Step 2: Apply the definition of the derivative
The derivative \( f'(x) \) is given by the limit:
\[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \]

Substituting the expressions for \( f(x + h) \) and \( f(x) \):
\[ f'(x) = \lim_{h \to 0} \frac{\frac{ax + ah + b}{ax + ah + d} - \frac{ax + b}{ax + d}}{h} \]

### Step 3: Simplify the expression
To simplify, we need a common denominator:
\[ f'(x) = \lim_{h \to 0} \frac{(ax + ah + b)(ax + d) - (ax + b)(ax + ah + d)}{h \cdot (ax + ah + d)(ax + d)} \]

### Step 4: Expand the numerators
Expanding both products in the numerator:
1. \( (ax + ah + b)(ax + d) = ax(ax + d) + ah(ax + d) + b(ax + d) \)
2. \( (ax + b)(ax + ah + d) = ax(ax + ah + d) + b(ax + ah + d) \)

After expanding and combining like terms, we will have:
\[ (ax^2 + a^2xh + adx + ahx + ahd + b(ax + d)) - (ax^2 + a^2xh + b(ax + ah + d)) \]

### Step 5: Cancel out terms
After canceling out the common terms, we will be left with:
\[ \lim_{h \to 0} \frac{(ad - ab)}{h \cdot (ax + h + d)(ax + d)} \]

### Step 6: Factor out \( h \) and take the limit
Now we can factor out \( h \):
\[ \lim_{h \to 0} \frac{(ad - ab)}{(ax + h + d)(ax + d)} \]

As \( h \) approaches 0, the expression simplifies to:
\[ \frac{ad - ab}{(ax + d)^2} \]

### Final Result
Thus, the derivative of \( f(x) \) is:
\[ f'(x) = \frac{ad - ab}{(ax + d)^2} \]