To differentiate the function $h(x) = \arctan(x \sin(x))$, we need to apply both the chain rule and the product rule. Here's a step-by-step explanation:

Step 1: Differentiate the outer function (arctan)

The derivative of $\arctan(u)$ with respect to $u$ is: $\frac{d}{du}[\arctan(u)] = \frac{1}{1 + u^2}$ So, for $h(x) = \arctan(x \sin(x))$, the derivative will be: $h'(x) = \frac{1}{1 + (x \sin(x))^2} \cdot \frac{d}{dx}[x \sin(x)]$

Step 2: Differentiate the inner function $x \sin(x)$

Now, using the product rule to differentiate $x \sin(x)$: $\frac{d}{dx}[x \sin(x)] = \sin(x) + x \cos(x)$

Step 3: Combine the results

Now, substitute this back into the derivative: $h'(x) = \frac{\sin(x) + x \cos(x)}{1 + (x \sin(x))^2}$

So, the derivative of $h(x) = \arctan(x \sin(x))$ is: $h'(x) = \frac{\sin(x) + x \cos(x)}{1 + (x \sin(x))^2}$

Would you like further details or clarification on this?

Here are 5 related questions for practice:

1. What is the derivative of $h(x) = \arctan(2x)$?
2. How would you differentiate $h(x) = \arcsin(x \sin(x))$?
3. Can you differentiate $h(x) = \arctan(\cos(x))$?
4. What is the second derivative of $h(x) = \arctan(x^2)$?
5. How would you find the derivative of $h(x) = \arctan(e^x)$?

Tip: When differentiating compositions like $\arctan(f(x))$, always remember to apply the chain rule carefully!