Use delta method to find the derivatives of the following :
`tan2x`

Answer

To find the derivative of the function \( y = \tan(2x) \) using the delta method, we follow these steps:

### Step 1: Define the function and the increment
Let:
\[ y = \tan(2x) \]
We introduce a small increment \( \delta x \) in \( x \), which gives us:
\[ y + \delta y = \tan(2(x + \delta x)) = \tan(2x + 2\delta x) \]

### Step 2: Express \( \delta y \)
From the definition, we can express \( \delta y \) as:
\[ \delta y = \tan(2x + 2\delta x) - \tan(2x) \]

### Step 3: Use the tangent addition formula
Using the tangent addition formula:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
we can substitute \( A = 2x \) and \( B = 2\delta x \):
\[ \tan(2x + 2\delta x) = \frac{\tan(2x) + \tan(2\delta x)}{1 - \tan(2x) \tan(2\delta x)} \]

### Step 4: Substitute back into \( \delta y \)
Now substituting this back into our expression for \( \delta y \):
\[ \delta y = \frac{\tan(2x) + \tan(2\delta x)}{1 - \tan(2x) \tan(2\delta x)} - \tan(2x) \]

### Step 5: Simplify \( \delta y \)
To simplify \( \delta y \), we can combine the terms:
\[ \delta y = \frac{\tan(2x) + \tan(2\delta x) - \tan(2x)(1 - \tan(2x) \tan(2\delta x)}{1 - \tan(2x) \tan(2\delta x)} \]
This simplifies to:
\[ \delta y = \frac{\tan(2\delta x)}{1 - \tan(2x) \tan(2\delta x)} \]

### Step 6: Divide by \( \delta x \)
Now, we divide both sides by \( \delta x \):
\[ \frac{\delta y}{\delta x} = \frac{\tan(2\delta x)}{\delta x(1 - \tan(2x) \tan(2\delta x))} \]

### Step 7: Take the limit as \( \delta x \to 0 \)
We know that \( \lim_{\delta x \to 0} \frac{\tan(2\delta x)}{2\delta x} = 1 \). Thus:
\[ \lim_{\delta x \to 0} \frac{\delta y}{\delta x} = \lim_{\delta x \to 0} \frac{2}{1 - \tan(2x) \cdot 0} = \frac{2}{1} = 2 \]

### Step 8: Final derivative
Thus, the derivative of \( y = \tan(2x) \) is:
\[ \frac{dy}{dx} = 2 \sec^2(2x) \]