Use delta method to find the derivatives of the following :
`sec(2x-1)`

Answer

To find the derivative of the function \( f(x) = \sec(2x - 1) \) using the delta method, we can follow these steps:

### Step 1: Define the Derivative
The derivative of a function \( f(x) \) using the delta method is defined as:
\[ f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} \]

### Step 2: Substitute the Function
Substituting \( f(x) = \sec(2x - 1) \) into the derivative formula gives:
\[ f'(x) = \lim_{\Delta x \to 0} \frac{\sec(2(x + \Delta x) - 1) - \sec(2x - 1)}{\Delta x} \]

### Step 3: Simplify the Expression
We can simplify the expression inside the limit:
\[ f'(x) = \lim_{\Delta x \to 0} \frac{\sec(2x + 2\Delta x - 1) - \sec(2x - 1)}{\Delta x} \]

### Step 4: Use the Identity for Secant
Recall that \( \sec \theta = \frac{1}{\cos \theta} \). Thus, we can rewrite the secant terms:
\[ f'(x) = \lim_{\Delta x \to 0} \frac{\frac{1}{\cos(2x + 2\Delta x - 1)} - \frac{1}{\cos(2x - 1)}}{\Delta x} \]

### Step 5: Combine the Fractions
To combine the fractions, we find a common denominator:
\[ f'(x) = \lim_{\Delta x \to 0} \frac{\cos(2x - 1) - \cos(2x + 2\Delta x - 1)}{\Delta x \cdot \cos(2x + 2\Delta x - 1) \cdot \cos(2x - 1)} \]

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

### Step 7: Simplify Further
As \( \Delta x \to 0 \), \( \sin(\Delta x) \approx \Delta x \):
\[ f'(x) = \lim_{\Delta x \to 0} \frac{-2 \sin(2x - 1) \cdot \Delta x}{\Delta x \cdot \cos(2x + 2\Delta x - 1) \cdot \cos(2x - 1)} \]

### Step 8: Cancel Out \( \Delta x \)
Canceling \( \Delta x \):
\[ f'(x) = \lim_{\Delta x \to 0} \frac{-2 \sin(2x - 1)}{\cos(2x + 2\Delta x - 1) \cdot \cos(2x - 1)} \]

### Step 9: Evaluate the Limit
As \( \Delta x \to 0 \), \( \cos(2x + 2\Delta x - 1) \to \cos(2x - 1) \):
\[ f'(x) = \frac{-2 \sin(2x - 1)}{\cos^2(2x - 1)} \]

### Step 10: Final Result
Using the identity \( \tan \theta = \frac{\sin \theta}{\cos \theta} \):
\[ f'(x) = -2 \tan(2x - 1) \sec^2(2x - 1) \]

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