Answer: Approximate change \(=\left(\frac{2\pi}{3\sqrt3}+\frac{2\pi^2}{27}\right)h\approx1.94h\).

For

\(y=x^2\tan\frac{x}{2},\)

differentiate using the product rule:

\(\frac{\mathrm{d}y}{\mathrm{d}x}=2x\tan\frac{x}{2}+\frac{x^2}{2}\sec^2\frac{x}{2}.\)

For a small increase \(h\),

\(\delta y\approx h\left.\frac{\mathrm{d}y}{\mathrm{d}x}\right|_{x=\frac{\pi}{3}}.\)

At \(x=\frac{\pi}{3}\), \(\tan\frac{x}{2}=\tan\frac{\pi}{6}=\frac{1}{\sqrt3}\) and \(\sec^2\frac{\pi}{6}=\frac43\).

Therefore

\(\left.\frac{\mathrm{d}y}{\mathrm{d}x}\right|_{x=\frac{\pi}{3}}=\frac{2\pi}{3\sqrt3}+\frac{2\pi^2}{27}=1.94\ldots.\)

So the approximate change is

\(1.94h.\)