Find from first principles the differential coefficient of `sin 2x`.

Answer

To find the differential coefficient of \( \sin 2x \) from first principles, we will follow these steps:

### Step 1: Define the function
Let \( g(x) = \sin(2x) \).

### Step 2: Write the expression for the derivative
The derivative \( g'(x) \) can be defined using the limit definition of the derivative:
\[ g'(x) = \lim_{h \to 0} \frac{g(x+h) - g(x)}{h} \]

### Step 3: Substitute \( g(x) \) and \( g(x+h) \)
We need to find \( g(x+h) \):
\[ g(x+h) = \sin(2(x+h)) = \sin(2x + 2h) \]
Now substituting into the derivative formula:
\[ g'(x) = \lim_{h \to 0} \frac{\sin(2x + 2h) - \sin(2x)}{h} \]

### Step 4: Use the sine difference identity
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 \( A = 2x + 2h \) and \( B = 2x \):
\[ g'(x) = \lim_{h \to 0} \frac{2 \cos\left(\frac{(2x + 2h) + 2x}{2}\right) \sin\left(\frac{(2x + 2h) - 2x}{2}\right)}{h} \]
This simplifies to:
\[ g'(x) = \lim_{h \to 0} \frac{2 \cos(2x + h) \sin(h)}{h} \]

### Step 5: Simplify the limit
Now we can separate the limit:
\[ g'(x) = 2 \cos(2x + h) \cdot \lim_{h \to 0} \frac{\sin(h)}{h} \]
We know that \( \lim_{h \to 0} \frac{\sin(h)}{h} = 1 \), so:
\[ g'(x) = 2 \cos(2x + 0) \cdot 1 = 2 \cos(2x) \]

### Conclusion
Thus, the differential coefficient of \( \sin(2x) \) is:
\[ g'(x) = 2 \cos(2x) \]

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