Find the derivative of `sin x` at x=0

Answer

To find the derivative of \( \sin x \) at \( x = 0 \), we will use the definition of the derivative. The derivative of a function \( f(x) \) at a point \( a \) is given by:

\[ f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} \]

In this case, we have \( f(x) = \sin x \) and we want to find \( f'(0) \).

### Step 1: Set up the limit for the derivative at \( x = 0 \)

Using the definition of the derivative, we can write:

\[ f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x - 0} \]

### Step 2: Substitute the function and the value at \( x = 0 \)

We know that \( f(0) = \sin(0) = 0 \). Thus, we can substitute this into our limit:

\[ f'(0) = \lim_{x \to 0} \frac{\sin x - 0}{x - 0} = \lim_{x \to 0} \frac{\sin x}{x} \]

### Step 3: Evaluate the limit

We need to evaluate the limit:

\[ \lim_{x \to 0} \frac{\sin x}{x} \]

It is a well-known limit that:

\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]

### Step 4: Conclude the derivative

Thus, we have:

\[ f'(0) = 1 \]

So, the derivative of \( \sin x \) at \( x = 0 \) is \( 1 \).

### Final Answer

The derivative of \( \sin x \) at \( x = 0 \) is:

\[ \boxed{1} \]