Differentiate the following functions w.r.t. x:
`x sin x`

Answer

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

\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]

In our case, let:
- \( u = x \)
- \( v = \sin x \)

Now, we will differentiate both \( u \) and \( v \):

1. **Differentiate \( u = x \)**:
\[ \frac{du}{dx} = 1 \]

2. **Differentiate \( v = \sin x \)**:
\[ \frac{dv}{dx} = \cos x \]

Now, applying the product rule:

\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]

Substituting the values of \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \):

\[ \frac{dy}{dx} = x \cdot \cos x + \sin x \cdot 1 \]

This simplifies to:

\[ \frac{dy}{dx} = x \cos x + \sin x \]

Thus, the derivative of \( y = x \sin x \) with respect to \( x \) is:

\[ \frac{dy}{dx} = x \cos x + \sin x \]