Find derivative of given functions w.r.t. the independent variable x.
`x sin x`

Options

A. `x cos x + sin x`
B. `x sin x + cos x`
C. `x + cos x`
D. `1+cosx`

Answer

To find the derivative of 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 two functions multiplied together, say \( u \) 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 can identify:
- \( u = x \)
- \( v = \sin x \)

Now, we will differentiate each part:

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 we have:

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

This simplifies to:

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

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

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