PRODUCT RULE y = (3x2 + 5)(4x4 + 3)

Let $U = 3x^2 + 5$.
Then $\frac{dU}{dx} = \frac{d}{dx}(3x^2 + 5) = 3(2x^{2-1}) + 0 = 6x$.
This matches your calculation for $\frac{dU}{dx}$.

Let $V = 4x^4 + 3$.
Then $\frac{dV}{dx} = \frac{d}{dx}(4x^4 + 3) = 4(4x^{4-1}) + 0 = 16x^3$.
Correction: Your calculation for $\frac{dV}{dx}$ was $16x$, but it should be $16x^3$.

Using the correct derivatives:

$$\frac{dy}{dx} = (3x^2 + 5)(16x^3) + (4x^4 + 3)(6x)$$

Note: In your written work, you used $16x$ for $\frac{dV}{dx}$ and $4x^3+3$ for $V$ in this step. However, your subsequent expansion suggests you implicitly used the correct terms.

$$\frac{dy}{dx} = (3x^2)(16x^3) + (5)(16x^3) + (4x^4)(6x) + (3)(6x)$$ $$\frac{dy}{dx} = 48x^{2+3} + 80x^3 + 24x^{4+1} + 18x$$ $$\frac{dy}{dx} = 48x^5 + 80x^3 + 24x^5 + 18x$$

This expansion matches your work.

$$\frac{dy}{dx} = (48x^5 + 24x^5) + 80x^3 + 18x$$ $$\frac{dy}{dx} = 72x^5 + 80x^3 + 18x$$

This matches your final answer.

Despite the minor errors in writing down $\frac{dV}{dx}$ and $V$ in the intermediate steps, your final result is correct because you applied the correct terms during the expansion.

The final answer is $72x^5 + 80x^3 + 18x$.