Differentiate by `1^(st)`, principle, f(x) = sec (1 - 2x).

Answer

To differentiate the function \( f(x) = \sec(1 - 2x) \) using the first principle of differentiation, we follow these steps:

### Step 1: Write the definition of the derivative
The derivative of a function \( f(x) \) using the first principle is given by:
\[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \]

### Step 2: Substitute the function into the definition
Substituting \( f(x) = \sec(1 - 2x) \):
\[ f'(x) = \lim_{h \to 0} \frac{\sec(1 - 2(x + h)) - \sec(1 - 2x)}{h} \]

### Step 3: Simplify the expression
This can be rewritten as:
\[ f'(x) = \lim_{h \to 0} \frac{\sec(1 - 2x - 2h) - \sec(1 - 2x)}{h} \]

### Step 4: Use the identity for secant
Recall that \( \sec(x) = \frac{1}{\cos(x)} \). Thus, we can express the difference of secants:
\[ f'(x) = \lim_{h \to 0} \frac{\frac{1}{\cos(1 - 2x - 2h)} - \frac{1}{\cos(1 - 2x)}}{h} \]

### Step 5: Combine the fractions
Finding a common denominator gives:
\[ f'(x) = \lim_{h \to 0} \frac{\cos(1 - 2x) - \cos(1 - 2x - 2h)}{h \cdot \cos(1 - 2x) \cdot \cos(1 - 2x - 2h)} \]

### Step 6: Use the cosine difference identity
Using the identity \( \cos A - \cos B = -2 \sin\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right) \):
\[ A = 1 - 2x, \quad B = 1 - 2x - 2h \]
Thus:
\[ f'(x) = \lim_{h \to 0} \frac{-2 \sin\left(1 - 2x - h\right) \sin(h)}{h \cdot \cos(1 - 2x) \cdot \cos(1 - 2x - 2h)} \]

### Step 7: Simplify the limit
As \( h \to 0 \), \( \sin(h) \approx h \):
\[ f'(x) = \lim_{h \to 0} \frac{-2 \sin(1 - 2x) \cdot h}{h \cdot \cos(1 - 2x) \cdot \cos(1 - 2x)} \]
This simplifies to:
\[ f'(x) = \frac{-2 \sin(1 - 2x)}{\cos^2(1 - 2x)} \]

### Step 8: Final result
Using the identity \( \tan(x) = \frac{\sin(x)}{\cos(x)} \):
\[ f'(x) = -2 \tan(1 - 2x) \sec(1 - 2x) \]

### Conclusion
Thus, the derivative of \( f(x) = \sec(1 - 2x) \) is:
\[ f'(x) = -2 \tan(1 - 2x) \sec(1 - 2x) \]