Let's solve for the second derivative $\frac{d^2y}{dx^2}$ for the function $y = 8x \sin(x^2)$.

Step 1: First derivative $\frac{dy}{dx}$

We need to apply the product rule to differentiate $y = 8x \sin(x^2)$. The product rule states that if you have two functions $u(x)$ and $v(x)$, then:

$\frac{d}{dx}[u(x) v(x)] = u'(x)v(x) + u(x)v'(x)$

Here, we set:

• $u(x) = 8x$
• $v(x) = \sin(x^2)$

Let's differentiate each term:

• $u'(x) = \frac{d}{dx}(8x) = 8$
• For $v'(x) = \frac{d}{dx}[\sin(x^2)]$, we use the chain rule: $v'(x) = \cos(x^2) \cdot \frac{d}{dx}(x^2) = \cos(x^2) \cdot 2x$

Now applying the product rule:

$\frac{dy}{dx} = 8 \sin(x^2) + 8x \cdot (2x \cos(x^2))$

Simplifying:

$\frac{dy}{dx} = 8 \sin(x^2) + 16x^2 \cos(x^2)$

Step 2: Second derivative $\frac{d^2y}{dx^2}$

Now we need to differentiate $\frac{dy}{dx} = 8 \sin(x^2) + 16x^2 \cos(x^2)$.

Again, use the product rule for both terms.

• Differentiate $8 \sin(x^2)$: $\frac{d}{dx}[8 \sin(x^2)] = 8 \cdot \cos(x^2) \cdot 2x = 16x \cos(x^2)$

• Differentiate $16x^2 \cos(x^2)$: Applying the product rule again: $\frac{d}{dx}[16x^2 \cos(x^2)] = 32x \cos(x^2) + 16x^2 (-\sin(x^2)) \cdot 2x$ Simplifying: $32x \cos(x^2) - 32x^3 \sin(x^2)$

Now, sum the derivatives:

$\frac{d^2y}{dx^2} = 16x \cos(x^2) + 32x \cos(x^2) - 32x^3 \sin(x^2)$

Simplifying:

$\frac{d^2y}{dx^2} = 48x \cos(x^2) - 32x^3 \sin(x^2)$

Final Answer:

$\frac{d^2y}{dx^2} = 48x \cos(x^2) - 32x^3 \sin(x^2)$

Would you like further details or have any questions?

Here are 5 related questions to explore:

1. How do you apply the chain rule to more complex trigonometric functions?
2. What is the product rule, and how is it useful in differentiation?
3. How would the result change if the function involved different trigonometric functions?
4. What is the importance of simplifying the second derivative?
5. How do you find higher-order derivatives for trigonometric functions?

Tip: Always double-check the application of both the chain and product rules in problems that involve multiple functions.