Differentiate the following with respect to x using first principle method.
`"cosec"(2x+3)`

Answer

To differentiate the function \( f(x) = \csc(2x + 3) \) using the first principle method, we follow these steps:

### Step 1: Write the definition of the derivative using the first principle
The derivative of a function \( f(x) \) using the first principle is given by:
\[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \]
For our function, we have:
\[ f'(x) = \lim_{h \to 0} \frac{\csc(2(x + h) + 3) - \csc(2x + 3)}{h} \]

### Step 2: Simplify the expression inside the limit
We can rewrite \( f(x + h) \):
\[ f(x + h) = \csc(2(x + h) + 3) = \csc(2x + 2h + 3) \]
Thus, our expression becomes:
\[ f'(x) = \lim_{h \to 0} \frac{\csc(2x + 2h + 3) - \csc(2x + 3)}{h} \]

### Step 3: Use the identity for cosecant
Recall that \( \csc(\theta) = \frac{1}{\sin(\theta)} \). Therefore, we can rewrite the limit as:
\[ f'(x) = \lim_{h \to 0} \frac{\frac{1}{\sin(2x + 2h + 3)} - \frac{1}{\sin(2x + 3)}}{h} \]

### Step 4: Combine the fractions
To combine the fractions, we find a common denominator:
\[ f'(x) = \lim_{h \to 0} \frac{\sin(2x + 3) - \sin(2x + 2h + 3)}{h \cdot \sin(2x + 3) \cdot \sin(2x + 2h + 3)} \]

### Step 5: Use the sine subtraction formula
Using the sine subtraction formula, we have:
\[ \sin A - \sin B = 2 \cos\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right) \]
Let \( A = 2x + 3 \) and \( B = 2x + 2h + 3 \):
\[ f'(x) = \lim_{h \to 0} \frac{2 \cos\left(2x + 3 + h\right) \sin\left(-h\right)}{h \cdot \sin(2x + 3) \cdot \sin(2x + 2h + 3)} \]

### Step 6: Simplify the limit
Since \( \sin(-h) = -\sin(h) \), we can rewrite the limit:
\[ f'(x) = \lim_{h \to 0} \frac{-2 \cos(2x + 3 + h) \sin(h)}{h \cdot \sin(2x + 3) \cdot \sin(2x + 2h + 3)} \]
Using the fact that \( \lim_{h \to 0} \frac{\sin(h)}{h} = 1 \):
\[ f'(x) = -2 \cdot \frac{\cos(2x + 3)}{\sin(2x + 3) \cdot \sin(2x + 3)} \]

### Step 7: Final expression
Thus, we have:
\[ f'(x) = -2 \cdot \frac{\cos(2x + 3)}{\sin^2(2x + 3)} \]
This can also be expressed as:
\[ f'(x) = -2 \cot(2x + 3) \csc(2x + 3) \]