Differentiate w.r.t. 'x' from first principles f (x) `= sqrt(4 - x)`

Answer

To differentiate the function \( f(x) = \sqrt{4 - x} \) with respect to \( x \) from first principles, we will follow these steps:

### Step 1: Write the definition of the derivative
The derivative of a function \( f(x) \) at a point \( x \) can be defined as:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

### Step 2: Substitute \( f(x) \) into the definition
Given \( f(x) = \sqrt{4 - x} \), we need to find \( f(x+h) \):
\[ f(x+h) = \sqrt{4 - (x + h)} = \sqrt{4 - x - h} \]
Now, substituting \( f(x) \) and \( f(x+h) \) into the derivative formula:
\[ f'(x) = \lim_{h \to 0} \frac{\sqrt{4 - x - h} - \sqrt{4 - x}}{h} \]

### Step 3: Rationalize the numerator
To simplify the expression, we will 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 4: 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 \( h \) from the numerator and denominator gives:
\[ f'(x) = \lim_{h \to 0} \frac{-1}{\sqrt{4 - x - h} + \sqrt{4 - x}} \]

### Step 5: Evaluate the limit as \( h \to 0 \)
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 \( f(x) = \sqrt{4 - x} \) is:
\[ f'(x) = \frac{-1}{2\sqrt{4 - x}} \]