Use delta method to find the derivatives of the following :
`xsinx`

Answer

To find the derivative of the function \( y = x \sin x \) using the delta method, we will follow these steps:

### Step 1: Define the function and increment
Let \( y = x \sin x \). We will consider an increment \( \Delta x \) in \( x \), which gives us a new value \( x + \Delta x \). The corresponding value of \( y \) will be \( y + \Delta y \).

### Step 2: Express \( y + \Delta y \)
We can express \( y + \Delta y \) as:
\[ y + \Delta y = (x + \Delta x) \sin(x + \Delta x) \]

### Step 3: Expand \( \sin(x + \Delta x) \)
Using the angle addition formula for sine, we have:
\[ \sin(x + \Delta x) = \sin x \cos(\Delta x) + \cos x \sin(\Delta x) \]
Thus,
\[ y + \Delta y = (x + \Delta x)(\sin x \cos(\Delta x) + \cos x \sin(\Delta x)) \]

### Step 4: Substitute and simplify
Substituting this back, we get:
\[ \Delta y = (x + \Delta x)(\sin x \cos(\Delta x) + \cos x \sin(\Delta x)) - x \sin x \]
Expanding this gives:
\[ \Delta y = x \sin x \cos(\Delta x) + x \cos x \sin(\Delta x) + \Delta x \sin x \cos(\Delta x) + \Delta x \cos x \sin(\Delta x) - x \sin x \]
This simplifies to:
\[ \Delta y = x \sin x (\cos(\Delta x) - 1) + x \cos x \sin(\Delta x) + \Delta x \sin x \cos(\Delta x) + \Delta x \cos x \sin(\Delta x) \]

### Step 5: Divide by \( \Delta x \)
Now, we divide \( \Delta y \) by \( \Delta x \):
\[ \frac{\Delta y}{\Delta x} = \frac{x \sin x (\cos(\Delta x) - 1)}{\Delta x} + \frac{x \cos x \sin(\Delta x)}{\Delta x} + \sin x \cos(\Delta x) + \cos x \sin(\Delta x) \]

### Step 6: Take the limit as \( \Delta x \to 0 \)
Now, we take the limit as \( \Delta x \to 0 \):
1. As \( \Delta x \to 0 \), \( \frac{\cos(\Delta x) - 1}{\Delta x} \to 0 \).
2. \( \frac{\sin(\Delta x)}{\Delta x} \to 1 \).

Thus, we have:
\[ \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = 0 + x \cos x + \sin x + 0 = x \cos x + \sin x \]

### Final Result
Therefore, the derivative of \( y = x \sin x \) is:
\[ \frac{dy}{dx} = x \cos x + \sin x \]

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