Differentiate `Sin^(2)x` with respect to x using first principle method.

Answer

To differentiate \( \sin^2 x \) with respect to \( x \) using the first principle method, we follow these steps:

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

### Step 2: Apply the first principle of derivatives
According to the first principle of derivatives, the derivative \( f'(x) \) is given by:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

### Step 3: Substitute \( f(x) \) and \( f(x+h) \)
Now, we need to find \( f(x+h) \):
\[ f(x+h) = \sin^2(x+h) \]
Thus, we can write:
\[ f'(x) = \lim_{h \to 0} \frac{\sin^2(x+h) - \sin^2 x}{h} \]

### Step 4: Use the identity for the difference of squares
We can use the identity \( a^2 - b^2 = (a-b)(a+b) \):
\[ \sin^2(x+h) - \sin^2 x = (\sin(x+h) - \sin x)(\sin(x+h) + \sin x) \]
So we rewrite the derivative:
\[ f'(x) = \lim_{h \to 0} \frac{(\sin(x+h) - \sin x)(\sin(x+h) + \sin x)}{h} \]

### Step 5: Break it into two limits
We know from the limit definition of the derivative that:
\[ \lim_{h \to 0} \frac{\sin(x+h) - \sin x}{h} = \cos x \]
Thus, we can express our limit as:
\[ f'(x) = \lim_{h \to 0} \left( \sin(x+h) + \sin x \right) \cdot \lim_{h \to 0} \frac{\sin(x+h) - \sin x}{h} \]
As \( h \to 0 \), \( \sin(x+h) \to \sin x \):
\[ f'(x) = \lim_{h \to 0} \left( \sin(x+h) + \sin x \right) \cdot \cos x = 2 \sin x \cdot \cos x \]

### Step 6: Final result
Thus, the derivative of \( \sin^2 x \) is:
\[ f'(x) = 2 \sin x \cos x \]
This can also be expressed using the double angle formula:
\[ f'(x) = \sin(2x) \]