The given function is:
[tex]y=5(x^2-4x)[/tex]
Therefore,
[tex]\begin{gathered} \frac{dy}{dx}=5(2x-4) \\ \frac{dy}{dx}=10(x-2) \end{gathered}[/tex]
Recall that:
[tex]\begin{gathered} \frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt} \\ \text{ Substitute }\frac{dy}{dx}=10(x-2)\text{ into the equation:} \\ \frac{dy}{dt}=10\frac{dx}{dt}(x-2) \end{gathered}[/tex]
Substitute x = 2 and dx/dt = 2
[tex]\frac{dy}{dt}=10(2-2)\cdot2=0[/tex]
Next, Substitute x = 4 and dy/dt = 8 into the equation:
[tex]\begin{gathered} 8=10\frac{dx}{dt}(4-2) \\ 20\frac{dx}{dt}=8 \\ \frac{dx}{dt}=\frac{2}{5} \end{gathered}[/tex]
(a) dy/dt = 0
(b) dx/dt = 2/5
