To determine the second derivative of $h(x) = x\sin(x)$, we first need to find the first derivative $h'(x)$ and then the second derivative $h''(x)$.

Step 1: Find the First Derivative

Using the product rule, which states $(uv)' = u'v + uv'$, we can differentiate $h(x) = x\sin(x)$.

Let $u = x$ and $v = \sin(x)$.

Then, $u' = 1$ $v' = \cos(x)$

So, $h'(x) = u'v + uv' = 1 \cdot \sin(x) + x \cdot \cos(x)$ $h'(x) = \sin(x) + x\cos(x)$

Step 2: Find the Second Derivative

Now, we differentiate $h'(x) = \sin(x) + x\cos(x)$ to find $h''(x)$.

Using the product rule again on $x\cos(x)$, $\left( x\cos(x) \right)' = x' \cos(x) + x \left( \cos(x) \right)'$ $= 1 \cdot \cos(x) + x \left( -\sin(x) \right)$ $= \cos(x) - x\sin(x)$

So, $h''(x) = \left( \sin(x) + x\cos(x) \right)'$ $= \left( \sin(x) \right)' + \left( x\cos(x) \right)'$ $= \cos(x) + \left( \cos(x) - x\sin(x) \right)$ $= \cos(x) + \cos(x) - x\sin(x)$ $= 2\cos(x) - x\sin(x)$

Therefore, the second derivative of $h(x) = x\sin(x)$ is: $h''(x) = 2\cos(x) - x\sin(x)$

Would you like more details or have any questions?

Here are 5 questions you might ask:

1. How do you apply the product rule for derivatives?
2. Can you explain the chain rule for derivatives?
3. How do you find the derivative of a trigonometric function?
4. What are common mistakes to avoid when differentiating products of functions?
5. Can you show an example of finding a second derivative for a different function?

Tip: Always carefully apply the product and chain rules when differentiating products or compositions of functions.