Differentiate the following w.r.t. x
`x*sinx*e^(x)`

Answer

To differentiate the function \( y = x \sin x \cdot e^x \) with respect to \( x \), we will use the product rule of differentiation. The product rule states that if you have two functions \( u \) and \( v \), then the derivative of their product is given by:

\[ \frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx} \]

In our case, we have three functions multiplied together: \( u = x \sin x \), \( v = e^x \). We will first differentiate \( y \) with respect to \( x \).

### Step 1: Identify the functions
Let:
- \( u = x \sin x \)
- \( v = e^x \)

### Step 2: Differentiate \( v \)
The derivative of \( v \) is:
\[ \frac{dv}{dx} = \frac{d}{dx}(e^x) = e^x \]

### Step 3: Differentiate \( u \)
Now, we need to differentiate \( u = x \sin x \). Here, we will again apply the product rule:
Let:
- \( a = x \)
- \( b = \sin x \)

Then,
\[ \frac{du}{dx} = a \frac{db}{dx} + b \frac{da}{dx} \]
Calculating the derivatives:
\[ \frac{da}{dx} = 1 \quad \text{and} \quad \frac{db}{dx} = \cos x \]
Thus,
\[ \frac{du}{dx} = x \cos x + \sin x \]

### Step 4: Apply the product rule for \( y \)
Now we can apply the product rule to find \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]
Substituting the values we have:
\[ \frac{dy}{dx} = (x \sin x) e^x + e^x (x \cos x + \sin x) \]

### Step 5: Factor out \( e^x \)
We can factor out \( e^x \):
\[ \frac{dy}{dx} = e^x \left( x \sin x + x \cos x + \sin x \right) \]

### Final Answer
Thus, the derivative of \( y = x \sin x \cdot e^x \) with respect to \( x \) is:
\[ \frac{dy}{dx} = e^x \left( x \sin x + x \cos x + \sin x \right) \]