Differentiate the function `"sin"(2x-3)` by First Principle of differentiation.

Answer

To differentiate the function \( f(x) = \sin(2x - 3) \) using the first principle of differentiation, we follow these steps:

### Step 1: Write the definition of the derivative
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 definition
Substituting \( f(x) = \sin(2x - 3) \):
\[ f'(x) = \lim_{h \to 0} \frac{\sin(2(x + h) - 3) - \sin(2x - 3)}{h} \]
This simplifies to:
\[ f'(x) = \lim_{h \to 0} \frac{\sin(2x + 2h - 3) - \sin(2x - 3)}{h} \]

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

### Step 4: Simplify the expression
Calculating the averages:
\[ \frac{(2x + 2h - 3) + (2x - 3)}{2} = 2x - 3 + h \]
\[ \frac{(2x + 2h - 3) - (2x - 3)}{2} = h \]
Thus, we have:
\[ f'(x) = \lim_{h \to 0} \frac{2 \cos(2x - 3 + h) \sin(h)}{h} \]

### Step 5: Split the limit
Now we can split the limit:
\[ f'(x) = \lim_{h \to 0} 2 \cos(2x - 3 + h) \cdot \lim_{h \to 0} \frac{\sin(h)}{h} \]

### Step 6: Evaluate the limits
We know that:
\[ \lim_{h \to 0} \frac{\sin(h)}{h} = 1 \]
And as \( h \to 0 \), \( \cos(2x - 3 + h) \to \cos(2x - 3) \):
\[ f'(x) = 2 \cos(2x - 3) \cdot 1 = 2 \cos(2x - 3) \]

### Final Answer
Thus, the derivative of the function \( f(x) = \sin(2x - 3) \) is:
\[ f'(x) = 2 \cos(2x - 3) \]