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

Answer

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

### Step 1: Write the definition of the derivative using first principles
The derivative of a function \( f(x) \) using the first principle is given by:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

### Step 2: Substitute the function into the definition
For our function \( f(x) = \sin(x^2) \), we need to find \( f(x+h) \):
\[ f(x+h) = \sin((x+h)^2) = \sin(x^2 + 2xh + h^2) \]
Now, substituting into the derivative formula:
\[ f'(x) = \lim_{h \to 0} \frac{\sin((x+h)^2) - \sin(x^2)}{h} \]

### Step 3: Apply the sine difference identity
We can use the sine difference identity:
\[ \sin A - \sin B = 2 \sin\left(\frac{A-B}{2}\right) \cos\left(\frac{A+B}{2}\right) \]
Let \( A = (x+h)^2 \) and \( B = x^2 \):
\[ A - B = (x^2 + 2xh + h^2) - x^2 = 2xh + h^2 \]
Thus,
\[ f'(x) = \lim_{h \to 0} \frac{2 \sin\left(\frac{2xh + h^2}{2}\right) \cos\left(\frac{(x+h)^2 + x^2}{2}\right)}{h} \]

### Step 4: Simplify the limit
Now, we can simplify:
\[ f'(x) = \lim_{h \to 0} \frac{2 \sin(xh + \frac{h^2}{2}) \cos\left(\frac{(x+h)^2 + x^2}{2}\right)}{h} \]
Factoring out \( h \) from the sine term:
\[ = \lim_{h \to 0} 2 \cos\left(\frac{(x+h)^2 + x^2}{2}\right) \cdot \frac{\sin(xh + \frac{h^2}{2})}{h} \]

### Step 5: Evaluate the limit
As \( h \to 0 \), \( \frac{\sin(xh + \frac{h^2}{2})}{h} \) approaches \( x \) because \( \sin(t) \approx t \) for small \( t \):
\[ \lim_{h \to 0} \frac{\sin(xh + \frac{h^2}{2})}{h} = x \]
Thus, we have:
\[ f'(x) = 2x \cdot \cos\left(\frac{x^2 + x^2}{2}\right) = 2x \cdot \cos(x^2) \]

### Final Result
The derivative of \( \sin(x^2) \) with respect to \( x \) is:
\[ f'(x) = 2x \cos(x^2) \]

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