Resolve the first one It's calculus Itiboribo

Let $u = 2x^3+4$.
Let $v = (5x-6)^2$.

Find the derivative of $u$:

$$u' = \frac{d}{dx}(2x^3+4) = 6x^2$$

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

$$v' = \frac{d}{dx}((5x-6)^2)$$

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

$$v' = \frac{dv}{dw} \cdot \frac{dw}{dx} = 2(5x-6) \cdot 5 = 10(5x-6)$$

The quotient rule states that if $y = \frac{u}{v}$, then $\frac{dy}{dx} = \frac{u'v - uv'}{v^2}$.
Substitute $u, v, u', v'$ into the formula:

$$\frac{dy}{dx} = \frac{(6x^2)(5x-6)^2 - (2x^3+4)(10(5x-6))}{((5x-6)^2)^2}$$ $$\frac{dy}{dx} = \frac{6x^2(5x-6)^2 - 10(2x^3+4)(5x-6)}{(5x-6)^4}$$

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

$$\frac{dy}{dx} = \frac{(5x-6)[6x^2(5x-6) - 10(2x^3+4)]}{(5x-6)^4}$$

Cancel one $(5x-6)$ term from the numerator and denominator:

$$\frac{dy}{dx} = \frac{6x^2(5x-6) - 10(2x^3+4)}{(5x-6)^3}$$

Expand the terms in the numerator:

$$\frac{dy}{dx} = \frac{30x^3 - 36x^2 - 20x^3 - 40}{(5x-6)^3}$$

Combine like terms in the numerator:

$$\frac{dy}{dx} = \frac{(30x^3 - 20x^3) - 36x^2 - 40}{(5x-6)^3}$$ $$\frac{dy}{dx} = \boxed{\frac{10x^3 - 36x^2 - 40}{(5x-6)^3}}$$