We are given the function $f(x) = \sin(x^2)$ and asked to find its derivative $f'(x)$.

To solve this, we will apply the chain rule. The chain rule is used when differentiating composite functions, and it states:

$\frac{d}{dx} \left[ \sin(g(x)) \right] = \cos(g(x)) \cdot g'(x)$

Step-by-step solution:

1. Start with $f(x) = \sin(x^2)$.
2. Apply the chain rule:
   • The outer function is $\sin(u)$, where $u = x^2$.
   • The derivative of $\sin(u)$ is $\cos(u)$, so we get $\cos(x^2)$.
   • Now, multiply by the derivative of the inner function $x^2$, which is $2x$.

Thus, the derivative is: $f'(x) = 2x \cos(x^2)$

Would you like further clarification or details on this? Here are 5 related questions for deeper exploration:

1. How is the chain rule applied to other trigonometric functions?
2. What is the derivative of $f(x) = \cos(x^2)$?
3. How would you differentiate $f(x) = \sin(x^3)$?
4. What happens when differentiating $f(x) = \sin(x^2 + 1)$?
5. Can you explain the second derivative of $f(x) = \sin(x^2)$?

Tip: When differentiating composite functions, always identify the outer and inner functions to apply the chain rule correctly.