Answer: \(\frac{\mathrm{d}y}{\mathrm{d}x}=\ln\frac12\) at \(x=0\), and the approximate change is \(h\ln\frac12\).

Use the quotient rule with

\(u=\ln(3x^2+16),\quad v=x+2.\)

Then

\(u'=\dfrac{6x}{3x^2+16},\quad v'=1.\)

Since \(y=\dfrac{u}{v}\),

\(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{v u'-u v'}{v^2}.\)

Therefore

\(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{(x+2)\dfrac{6x}{3x^2+16}-\ln(3x^2+16)}{(x+2)^2}.\)

At \(x=0\), this becomes

\(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{0-\ln16}{4}=-\dfrac{\ln16}{4}.\)

Since \(\ln16=\ln(2^4)=4\ln2\),

\(-\dfrac{\ln16}{4}=-\ln2=\ln\frac12.\)

For a small increase \(h\) from \(x=0\),

\(\Delta y\approx h\left.\dfrac{\mathrm{d}y}{\mathrm{d}x}\right|_{x=0}.\)

Hence the approximate change in \(y\) is

\(h\ln\frac12.\)