Differentiate. $$ y=\left(9 x^{2}+8\right)(5 x+4) $$ $$ y^{\prime}=\square $$

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We are given 
$$
y = \left(9x^2+8\right)(5x+4).
$$

To differentiate, we use the product rule which states that 
$$
y' = u'v + uv',
$$
where 
$$
u = 9x^2+8 \quad \text{and} \quad v = 5x+4.
$$

1. Differentiate $$ u $$:
   $$
   u' = \frac{d}{dx}(9x^2+8)=18x.
   $$

2. Differentiate $$ v $$:
   $$
   v' = \frac{d}{dx}(5x+4)=5.
   $$

3. Apply the product rule:
   $$
   y' = (18x)(5x+4) + (9x^2+8)(5).
   $$

4. Expand and simplify:
   $$
   (18x)(5x+4) = 90x^2 + 72x,
   $$
   and
   $$
   (9x^2+8)(5) = 45x^2 + 40.
   $$
   Therefore,
   $$
   y' = 90x^2 + 72x + 45x^2 + 40 = 135x^2 + 72x + 40.
   $$

Thus, the derivative is 
$$
y' = 135x^2 + 72x + 40.
$$
$$ y' = 135x^2 + 72x + 40 $$

Differentiate.

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