Use delta method to find the derivatives of the following :
`tan(3x+1)`

Answer

To find the derivative of the function \( y = \tan(3x + 1) \) using the delta method, we will follow these steps:

### Step 1: Define the function and increment
Let \( y = \tan(3x + 1) \). We will consider a small increment \( \Delta x \) in \( x \), which leads to a change in \( y \) denoted as \( \Delta y \).

### Step 2: Express \( \Delta y \)
The change in \( y \) can be expressed as:
\[ \Delta y = \tan(3(x + \Delta x) + 1) - \tan(3x + 1) \]
This simplifies to:
\[ \Delta y = \tan(3x + 3\Delta x + 1) - \tan(3x + 1) \]

### Step 3: Use the tangent subtraction formula
We can use the identity for the difference of tangents:
\[ \tan A - \tan B = \frac{\sin(A - B)}{\cos A \cos B} \]
Let \( A = 3x + 3\Delta x + 1 \) and \( B = 3x + 1 \). Thus,
\[ \Delta y = \frac{\sin((3x + 3\Delta x + 1) - (3x + 1))}{\cos(3x + 3\Delta x + 1) \cos(3x + 1)} \]
This simplifies to:
\[ \Delta y = \frac{\sin(3\Delta x)}{\cos(3x + 3\Delta x + 1) \cos(3x + 1)} \]

### Step 4: Divide by \( \Delta x \)
Now, we divide \( \Delta y \) by \( \Delta x \):
\[ \frac{\Delta y}{\Delta x} = \frac{\sin(3\Delta x)}{\Delta x \cdot \cos(3x + 3\Delta x + 1) \cos(3x + 1)} \]

### Step 5: Take the limit as \( \Delta x \to 0 \)
To find the derivative \( \frac{dy}{dx} \), we take the limit as \( \Delta x \) approaches 0:
\[ \frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\sin(3\Delta x)}{\Delta x \cdot \cos(3x + 3\Delta x + 1) \cos(3x + 1)} \]

### Step 6: Use the limit identity
Using the limit identity \( \lim_{u \to 0} \frac{\sin u}{u} = 1 \), we rewrite the expression:
\[ \frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\sin(3\Delta x)}{3\Delta x} \cdot \frac{3}{\cos(3x + 3\Delta x + 1) \cos(3x + 1)} \]
This gives us:
\[ \frac{dy}{dx} = 3 \cdot \lim_{\Delta x \to 0} \frac{\sin(3\Delta x)}{3\Delta x} \cdot \frac{1}{\cos(3x + 1) \cos(3x + 3\Delta x + 1)} \]

### Step 7: Evaluate the limit
As \( \Delta x \to 0 \), \( \cos(3x + 3\Delta x + 1) \to \cos(3x + 1) \):
\[ \frac{dy}{dx} = 3 \cdot 1 \cdot \frac{1}{\cos^2(3x + 1)} = \frac{3}{\cos^2(3x + 1)} \]

### Final Result
Thus, the derivative of \( y = \tan(3x + 1) \) is:
\[ \frac{dy}{dx} = 3 \sec^2(3x + 1) \]