Answer: \(\displaystyle \frac{dy}{dt}=-0.08\).

For a trigonometric graph, identify the amplitude, midline and period first, then use these to locate the key points of the sketch.

Differentiate

\(y=2+\tan(1-x)\)

with respect to \(x\):

\(\frac{dy}{dx}=-\operatorname{sec}^2(1-x).\)

When \(y=3\),

\(3=2+\tan(1-x).\)

So

\(\tan(1-x)=1.\)

Since \(0\leq x\leq\frac{\pi}{2}\), the relevant solution is

\(1-x=\frac{\pi}{4}.\)

Thus

\(\operatorname{sec}^2(1-x)=\operatorname{sec}^2\frac{\pi}{4}=2.\)

Hence

\(\frac{dy}{dx}=-2.\)

Given that

\(\frac{dx}{dt}=0.04,\)

the chain rule gives

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

Therefore

\(\frac{dy}{dt}=(-2)(0.04)=-0.08.\)