Differentiation

Term by term: $5x^4$ gives $4 \times 5x^3 = 20x^3$; $-2x$ gives $-2$; the constant $7$ gives $0$.

Rewrite as $y = 2x^{-3}$ first. Then $2 \times (-3)x^{-4} = -6x^{-4}$, which is $-6$ over $x^4$.

$\dfrac{dy}{dx} = 2x - 5$, and at $x = 4$ that is $8 - 5 = 3$. Substitute into the derivative, not into $y$.

Chain rule. Outside: $5(3x+2)^4$. Derivative of the inside: $3$. Multiply: $5 \times 3 = 15$.

Quotient rule with $u = x^2 + 1$ and $v = x - 2$. Numerator: $(x-2)(2x) - (x^2+1)(1) = 2x^2 - 4x - x^2 - 1 = x^2 - 4x - 1$.

At rest means the velocity is zero. $v = \dfrac{ds}{dt} = 3t^2 - 12t + 9 = 3(t-1)(t-3)$, so $t = 1$ or $t = 3$.