Given the curve \(y = \frac{8}{x} + 2x\), we need to find \(\frac{dy}{dt}\) when \(x = 1\) and \(\frac{dx}{dt} = 0.04\).

First, differentiate \(y\) with respect to \(x\):

\(\frac{dy}{dx} = \frac{d}{dx} \left( \frac{8}{x} + 2x \right) = -\frac{8}{x^2} + 2\).

At point \(A\) where \(x = 1\), substitute \(x = 1\) into \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = -\frac{8}{1^2} + 2 = -8 + 2 = -6\).

Now, use the chain rule to find \(\frac{dy}{dt}\):

\(\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}\).

Substitute \(\frac{dy}{dx} = -6\) and \(\frac{dx}{dt} = 0.04\):

\(\frac{dy}{dt} = -6 \times 0.04 = -0.24\).