The problem asks to find the first and second derivatives of $y = \sin(x^2)$. Let's solve it step by step.

1. First Derivative $y'$

To differentiate $y = \sin(x^2)$, we use the chain rule. The outer function is $\sin(u)$, where $u = x^2$.

[ y' = \frac{d}{dx}[\sin(x^2)] = \cos(x^2) \cdot \frac{d}{dx}[x^2] ]

Now, differentiate $x^2$:

$\frac{d}{dx}[x^2] = 2x$

Thus, the first derivative is:

$y' = 2x \cos(x^2)$

2. Second Derivative $y''$

Now, we need to find the second derivative of $y' = 2x \cos(x^2)$. We'll use the product rule and chain rule again.

The product rule states:

$\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$

Let $f(x) = 2x$ and $g(x) = \cos(x^2)$.

• Differentiate $f(x) = 2x$: $f'(x) = 2$

• Now differentiate $g(x) = \cos(x^2)$ using the chain rule: $g'(x) = -\sin(x^2) \cdot \frac{d}{dx}[x^2] = -\sin(x^2) \cdot 2x$ So, $g'(x) = -2x \sin(x^2)$

Now apply the product rule:

$y'' = 2 \cdot \cos(x^2) + 2x \cdot (-2x \sin(x^2))$

Simplifying:

$y'' = 2 \cos(x^2) - 4x^2 \sin(x^2)$

Final Answer:

• First derivative: $y' = 2x \cos(x^2)$
• Second derivative: $y'' = 2 \cos(x^2) - 4x^2 \sin(x^2)$

Would you like more details or have any questions?

Here are some related questions to consider:

1. How does the chain rule work when differentiating composite functions?
2. What is the product rule, and how is it applied?
3. Can you use the second derivative to identify concavity or points of inflection for this function?
4. What other trigonometric functions can be differentiated using the chain rule?
5. How would you interpret the behavior of the function $y = \sin(x^2)$ graphically?

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