Find the derivative of sin(4x - 1) using first principle of derivative

Answer

To find the derivative of the function \( f(x) = \sin(4x - 1) \) using the first principle of derivatives, we will 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 the formula:
\[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \]

### Step 2: Substitute the function into the formula
For our function \( f(x) = \sin(4x - 1) \), we need to calculate \( f(x + h) \):
\[ f(x + h) = \sin(4(x + h) - 1) = \sin(4x + 4h - 1) \]
Now, substitute \( f(x + h) \) and \( f(x) \) into the derivative formula:
\[ f'(x) = \lim_{h \to 0} \frac{\sin(4x + 4h - 1) - \sin(4x - 1)}{h} \]

### Step 3: Apply the sine difference formula
We can use the sine difference identity:
\[ \sin A - \sin B = 2 \cos\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right) \]
Let \( A = 4x + 4h - 1 \) and \( B = 4x - 1 \). Then,
\[ A - B = (4x + 4h - 1) - (4x - 1) = 4h \]
\[ A + B = (4x + 4h - 1) + (4x - 1) = 8x + 4h - 2 \]
Now, substituting these into the sine difference formula gives:
\[ \sin(4x + 4h - 1) - \sin(4x - 1) = 2 \cos\left(\frac{8x + 4h - 2}{2}\right) \sin\left(\frac{4h}{2}\right) \]
This simplifies to:
\[ = 2 \cos(4x + 2h - 1) \sin(2h) \]

### Step 4: Substitute back into the limit
Now substitute this back into the limit:
\[ f'(x) = \lim_{h \to 0} \frac{2 \cos(4x + 2h - 1) \sin(2h)}{h} \]
We can rewrite \( \frac{\sin(2h)}{h} \) as \( 2 \frac{\sin(2h)}{2h} \):
\[ f'(x) = \lim_{h \to 0} 2 \cos(4x + 2h - 1) \cdot 2 \frac{\sin(2h)}{2h} \]

### Step 5: Evaluate the limit
As \( h \to 0 \), \( \frac{\sin(2h)}{2h} \to 1 \) and \( \cos(4x + 2h - 1) \to \cos(4x - 1) \):
\[ f'(x) = 2 \cdot 2 \cdot \cos(4x - 1) \cdot 1 = 4 \cos(4x - 1) \]

### Final Answer
Thus, the derivative of \( \sin(4x - 1) \) is:
\[ f'(x) = 4 \cos(4x - 1) \]