Find the derivative of `sin x`.

Options

A. `cos x`
B. `sec^(2) x`
C. `- cosec^(2) x`
D. `sec x. tanx`

Answer

To find the derivative of \( \sin x \) using the first principle of derivatives, we will follow these steps:

### Step 1: Define the derivative using the first principle
The derivative of a function \( f(x) \) at a point \( x \) is defined as:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

### Step 2: Substitute \( f(x) = \sin x \)
We will substitute \( f(x) \) with \( \sin x \):
\[ f'(x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin x}{h} \]

### Step 3: Use the sine addition formula
Using the sine addition formula \( \sin(a + b) = \sin a \cos b + \cos a \sin b \), we can expand \( \sin(x+h) \):
\[ \sin(x+h) = \sin x \cos h + \cos x \sin h \]
Substituting this back into our limit gives:
\[ f'(x) = \lim_{h \to 0} \frac{(\sin x \cos h + \cos x \sin h) - \sin x}{h} \]

### Step 4: Simplify the expression
Now we can simplify the expression:
\[ f'(x) = \lim_{h \to 0} \frac{\sin x \cos h - \sin x + \cos x \sin h}{h} \]
This can be rearranged to:
\[ f'(x) = \lim_{h \to 0} \left( \frac{\sin x (\cos h - 1)}{h} + \frac{\cos x \sin h}{h} \right) \]

### Step 5: Split the limit
We can split the limit into two parts:
\[ f'(x) = \lim_{h \to 0} \frac{\sin x (\cos h - 1)}{h} + \lim_{h \to 0} \frac{\cos x \sin h}{h} \]

### Step 6: Evaluate the limits
1. The first limit \( \lim_{h \to 0} \frac{\cos h - 1}{h} = 0 \) (as \( \cos h \) approaches 1 when \( h \) approaches 0).
2. The second limit \( \lim_{h \to 0} \frac{\sin h}{h} = 1 \).

Thus, we have:
\[ f'(x) = \sin x \cdot 0 + \cos x \cdot 1 \]
This simplifies to:
\[ f'(x) = \cos x \]

### Conclusion
The derivative of \( \sin x \) is:
\[ \frac{d}{dx}(\sin x) = \cos x \]