Find the derivative of `sqrt(4-x)` w.r.t. `x` using the first principle.

Explanation

To find the derivative of \( f(x) = \sqrt{4 - x} \) with respect to \( x \) using the first principle, we will follow these steps:

### Step 1: Define the function and the increment
Let:
\[ f(x) = \sqrt{4 - x} \]
Then, we define:
\[ f(x + h) = \sqrt{4 - (x + h)} = \sqrt{4 - x - h} \]

### Step 2: Write the formula for the derivative using the first principle
The derivative \( f'(x) \) can be expressed using the limit definition:
\[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \]

### Step 3: Substitute the function values into the limit
Substituting \( f(x + h) \) and \( f(x) \) into the limit gives:
\[ f'(x) = \lim_{h \to 0} \frac{\sqrt{4 - x - h} - \sqrt{4 - x}}{h} \]

### Step 4: Rationalize the numerator
To simplify the expression, we multiply the numerator and the denominator by the conjugate of the numerator:
\[ f'(x) = \lim_{h \to 0} \frac{\left(\sqrt{4 - x - h} - \sqrt{4 - x}\right) \left(\sqrt{4 - x - h} + \sqrt{4 - x}\right)}{h \left(\sqrt{4 - x - h} + \sqrt{4 - x}\right)} \]
This simplifies to:
\[ f'(x) = \lim_{h \to 0} \frac{(4 - x - h) - (4 - x)}{h \left(\sqrt{4 - x - h} + \sqrt{4 - x}\right)} \]

### Step 5: Simplify the expression
The numerator simplifies to:
\[ (4 - x - h) - (4 - x) = -h \]
Thus, we have:
\[ f'(x) = \lim_{h \to 0} \frac{-h}{h \left(\sqrt{4 - x - h} + \sqrt{4 - x}\right)} \]
Cancelling \( h \) in the numerator and denominator gives:
\[ f'(x) = \lim_{h \to 0} \frac{-1}{\sqrt{4 - x - h} + \sqrt{4 - x}} \]

### Step 6: Evaluate the limit
As \( h \) approaches 0, \( \sqrt{4 - x - h} \) approaches \( \sqrt{4 - x} \):
\[ f'(x) = \frac{-1}{\sqrt{4 - x} + \sqrt{4 - x}} = \frac{-1}{2\sqrt{4 - x}} \]

### Final Result
Thus, the derivative of \( f(x) = \sqrt{4 - x} \) with respect to \( x \) is:
\[ f'(x) = \frac{-1}{2\sqrt{4 - x}} \]

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