Find the derivative of cot (3x + 5) using first principle of derivatives

Answer

To find the derivative of \( f(x) = \cot(3x + 5) \) using the first principle of derivatives, we will follow these steps:

### Step 1: Write the formula for the derivative using the first principle.
The derivative of a function \( f(x) \) using the first principle is given by:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

### Step 2: Substitute the function into the formula.
Here, \( f(x) = \cot(3x + 5) \). Therefore, we have:
\[ f'(x) = \lim_{h \to 0} \frac{\cot(3(x+h) + 5) - \cot(3x + 5)}{h} \]

### Step 3: Simplify the expression inside the limit.
This simplifies to:
\[ f'(x) = \lim_{h \to 0} \frac{\cot(3x + 3h + 5) - \cot(3x + 5)}{h} \]

### Step 4: Use the identity for the difference of cotangents.
Using the identity \( \cot A - \cot B = \frac{\sin(B - A)}{\sin A \sin B} \), we can rewrite the numerator:
\[ \cot(3x + 3h + 5) - \cot(3x + 5) = \frac{\sin((3x + 5) - (3x + 3h + 5))}{\sin(3x + 3h + 5) \sin(3x + 5)} \]
This gives:
\[ = \frac{\sin(-3h)}{\sin(3x + 3h + 5) \sin(3x + 5)} \]

### Step 5: Substitute this back into the limit.
Now substituting this back into the limit, we have:
\[ f'(x) = \lim_{h \to 0} \frac{\sin(-3h)}{h \cdot \sin(3x + 3h + 5) \sin(3x + 5)} \]

### Step 6: Simplify using the limit property.
We know that \( \lim_{h \to 0} \frac{\sin(kh)}{h} = k \) for any constant \( k \). Here, \( k = -3 \):
\[ f'(x) = \lim_{h \to 0} \frac{-3 \cdot h}{h \cdot \sin(3x + 3h + 5) \sin(3x + 5)} = \frac{-3}{\sin(3x + 5) \cdot \sin(3x + 5)} \]

### Step 7: Evaluate the limit.
As \( h \to 0 \), \( \sin(3x + 3h + 5) \to \sin(3x + 5) \):
\[ f'(x) = -3 \cdot \frac{1}{\sin^2(3x + 5)} \]

### Step 8: Rewrite in terms of cosecant.
Since \( \frac{1}{\sin^2(3x + 5)} = \csc^2(3x + 5) \), we have:
\[ f'(x) = -3 \csc^2(3x + 5) \]

### Final Answer:
Thus, the derivative of \( \cot(3x + 5) \) is:
\[ f'(x) = -3 \csc^2(3x + 5) \]