So find the second derivative of that same question

Let $U = 10x^3 - 36x^2 - 40$.
Let $V = (5x-6)^3$.

Find the derivative of $U$:

$$U' = \frac{d}{dx}(10x^3 - 36x^2 - 40) = 30x^2 - 72x$$

Find the derivative of $V$ using the chain rule:

$$V' = \frac{d}{dx}((5x-6)^3)$$

Let $w = 5x-6$, so $V = w^3$. Then $\frac{dV}{dw} = 3w^2$ and $\frac{dw}{dx} = 5$.

$$V' = \frac{dV}{dw} \cdot \frac{dw}{dx} = 3(5x-6)^2 \cdot 5 = 15(5x-6)^2$$

The quotient rule states that if $y = \frac{U}{V}$, then $\frac{d^2y}{dx^2} = \frac{U'V - UV'}{V^2}$.
Substitute $U, V, U', V'$ into the formula:

$$\frac{d^2y}{dx^2} = \frac{(30x^2 - 72x)(5x-6)^3 - (10x^3 - 36x^2 - 40)(15(5x-6)^2)}{((5x-6)^3)^2}$$ $$\frac{d^2y}{dx^2} = \frac{(30x^2 - 72x)(5x-6)^3 - 15(10x^3 - 36x^2 - 40)(5x-6)^2}{(5x-6)^6}$$

Factor out $(5x-6)^2$ from the numerator:

$$\frac{d^2y}{dx^2} = \frac{(5x-6)^2[(30x^2 - 72x)(5x-6) - 15(10x^3 - 36x^2 - 40)]}{(5x-6)^6}$$

Cancel $(5x-6)^2$ from the numerator and denominator:

$$\frac{d^2y}{dx^2} = \frac{(30x^2 - 72x)(5x-6) - 15(10x^3 - 36x^2 - 40)}{(5x-6)^4}$$

Expand the first product:

$$(30x^2 - 72x)(5x-6) = 30x^2(5x) + 30x^2(-6) - 72x(5x) - 72x(-6)$$ $$= 150x^3 - 180x^2 - 360x^2 + 432x = 150x^3 - 540x^2 + 432x$$

Expand the second product:

$$-15(10x^3 - 36x^2 - 40) = -150x^3 + 540x^2 + 600$$

Combine these two expanded parts for the numerator:

$$(150x^3 - 540x^2 + 432x) + (-150x^3 + 540x^2 + 600)$$ $$= (150x^3 - 150x^3) + (-540x^2 + 540x^2) + 432x + 600$$ $$= 432x + 600$$

Factor out the common factor from the numerator (24):

$$432x + 600 = 24(18x + 25)$$