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

Answer

To find the derivative of the function \( y = \cot(2x + 1) \) using the delta method, we will follow these steps:

### Step 1: Define the function
Let \( f(x) = \cot(2x + 1) \).

### Step 2: Apply the delta method formula
The derivative \( \frac{dy}{dx} \) can be expressed using the limit definition:
\[ \frac{dy}{dx} = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \]

### Step 3: Substitute \( f(x + h) \)
Now, we need to find \( f(x + h) \):
\[ f(x + h) = \cot(2(x + h) + 1) = \cot(2x + 2h + 1) \]
So, we can rewrite the limit:
\[ \frac{dy}{dx} = \lim_{h \to 0} \frac{\cot(2x + 2h + 1) - \cot(2x + 1)}{h} \]

### Step 4: Use the cotangent subtraction formula
Using the identity for the difference of cotangents:
\[ \cot A - \cot B = \frac{\sin(B - A)}{\sin A \sin B} \]
where \( A = 2x + 2h + 1 \) and \( B = 2x + 1 \), we can express the difference:
\[ \cot(2x + 2h + 1) - \cot(2x + 1) = \frac{\sin((2x + 1) - (2x + 2h + 1))}{\sin(2x + 2h + 1) \sin(2x + 1)} = \frac{\sin(-2h)}{\sin(2x + 2h + 1) \sin(2x + 1)} \]

### Step 5: Substitute back into the limit
Now substituting this back into our limit:
\[ \frac{dy}{dx} = \lim_{h \to 0} \frac{\frac{\sin(-2h)}{\sin(2x + 2h + 1) \sin(2x + 1)}}{h} \]
This simplifies to:
\[ \frac{dy}{dx} = \lim_{h \to 0} \frac{\sin(-2h)}{h \cdot \sin(2x + 2h + 1) \sin(2x + 1)} \]

### Step 6: Evaluate the limit
Using the limit property \( \lim_{h \to 0} \frac{\sin(kh)}{h} = k \) for small \( h \):
\[ \lim_{h \to 0} \frac{\sin(-2h)}{h} = -2 \]
Thus, we have:
\[ \frac{dy}{dx} = -2 \cdot \frac{1}{\sin(2x + 1) \cdot \sin(2x + 1)} = -\frac{2}{\sin^2(2x + 1)} \]

### Step 7: Use the cotangent identity
Recall that \( \frac{1}{\sin^2(2x + 1)} = \csc^2(2x + 1) \), so we can express our final result as:
\[ \frac{dy}{dx} = -2 \csc^2(2x + 1) \]

### Final Answer
The derivative of \( \cot(2x + 1) \) is:
\[ \frac{dy}{dx} = -2 \csc^2(2x + 1) \]