To find the first and second derivatives of the function $g(x) = x \sin(x)$, we will use the product rule.

First Derivative $g'(x)$:

The product rule states: $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$ For $g(x) = x \sin(x)$, we have:

• $u(x) = x$, so $u'(x) = 1$
• $v(x) = \sin(x)$, so $v'(x) = \cos(x)$

Using the product rule: $g'(x) = u'(x)v(x) + u(x)v'(x)$ $g'(x) = 1 \cdot \sin(x) + x \cdot \cos(x)$ $g'(x) = \sin(x) + x \cos(x)$

Second Derivative $g''(x)$:

Now, let's differentiate $g'(x) = \sin(x) + x \cos(x)$.

We differentiate each term:

• The derivative of $\sin(x)$ is $\cos(x)$
• The derivative of $x \cos(x)$ requires the product rule again:
  • Let $u(x) = x$, so $u'(x) = 1$
  • Let $v(x) = \cos(x)$, so $v'(x) = -\sin(x)$

Using the product rule again: $\frac{d}{dx}[x \cos(x)] = 1 \cdot \cos(x) + x \cdot (-\sin(x))$ $\frac{d}{dx}[x \cos(x)] = \cos(x) - x \sin(x)$

Thus, the second derivative is: $g''(x) = \cos(x) + (\cos(x) - x \sin(x))$ $g''(x) = 2 \cos(x) - x \sin(x)$

Final Results:

$g'(x) = \sin(x) + x \cos(x)$ $g''(x) = 2 \cos(x) - x \sin(x)$

Would you like more details on any of these steps? Here are some related questions:

1. What is the product rule and when do we use it in differentiation?
2. How do we apply the product rule to more complex functions?
3. What is the chain rule, and how does it differ from the product rule?
4. How do you differentiate trigonometric functions like $\sin(x)$ and $\cos(x)$?
5. Can you provide examples of higher-order derivatives?

Tip: When dealing with products of functions, always check if the product rule is necessary, and remember that it simplifies the process of differentiating the product.