Differentiate by `1^(st)` principal `f(x) = tan (1-2x)`

Answer

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

### Step-by-Step Solution:

1. **Understand the First Principle of Derivatives**:
The derivative of a function \( f(x) \) using the first principle is given by:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

2. **Substitute \( f(x) \)**:
We know that \( f(x) = \tan(1 - 2x) \). Therefore, we need to find \( f(x+h) \):
\[ f(x+h) = \tan(1 - 2(x+h)) = \tan(1 - 2x - 2h) \]

3. **Set Up the Limit**:
Substitute \( f(x+h) \) and \( f(x) \) into the limit:
\[ f'(x) = \lim_{h \to 0} \frac{\tan(1 - 2x - 2h) - \tan(1 - 2x)}{h} \]

4. **Use the Tangent Difference Formula**:
We can use the identity for the difference of tangents:
\[ \tan A - \tan B = \frac{\sin(A - B)}{\cos A \cos B} \]
Here, let \( A = 1 - 2x - 2h \) and \( B = 1 - 2x \):
\[ f'(x) = \lim_{h \to 0} \frac{\sin((1 - 2x - 2h) - (1 - 2x))}{\cos(1 - 2x - 2h) \cos(1 - 2x) \cdot h} \]
Simplifying the sine term:
\[ = \lim_{h \to 0} \frac{\sin(-2h)}{\cos(1 - 2x - 2h) \cos(1 - 2x) \cdot h} \]

5. **Apply the Limit**:
We know that \( \sin(-2h) = -2\sin(h) \) and as \( h \to 0 \), \( \frac{\sin(h)}{h} \to 1 \):
\[ f'(x) = \lim_{h \to 0} \frac{-2\sin(h)}{\cos(1 - 2x - 2h) \cos(1 - 2x) \cdot h} = \lim_{h \to 0} \frac{-2}{\cos(1 - 2x - 2h) \cos(1 - 2x)} \cdot \frac{\sin(h)}{h} \]
Thus, as \( h \to 0 \):
\[ f'(x) = \frac{-2}{\cos^2(1 - 2x)} \cdot 1 \]

6. **Final Result**:
Therefore, the derivative of \( f(x) \) is:
\[ f'(x) = -2 \sec^2(1 - 2x) \]