Answer: \(\dfrac{dy}{dx}=\dfrac{(x+2)\dfrac{4x}{2x^2+1}-\ln(2x^2+1)}{(x+2)^2}\); approximate change \(=\dfrac{4-\ln3}{9}h\); \(\dfrac{dx}{dt}=9.31\).

We start with the main method. For a small change, use \(\Delta y\approx (dy/dx)\Delta x\) evaluated at the starting value.

Write

\(y=\frac{\ln(2x^2+1)}{x+2}.\)

Use the quotient rule. The derivative of \(\ln(2x^2+1)\) is

\(\frac{4x}{2x^2+1}.\)

Therefore

\(\frac{dy}{dx} =\frac{(x+2)\dfrac{4x}{2x^2+1}-\ln(2x^2+1)}{(x+2)^2}.\)

At \(x=1\),

\(\frac{dy}{dx} =\frac{3\cdot\frac43-\ln3}{9} =\frac{4-\ln3}{9}.\)

So, for a small increase from \(1\) to \(1+h\),

\(\delta y\approx\frac{4-\ln3}{9}h.\)

For part (c), use the chain rule:

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

When \(x=1\), the rate of change in \(y\) is \(3\), so

\(3=\frac{4-\ln3}{9}\frac{dx}{dt}.\)

Hence

\(\frac{dx}{dt}=\frac{27}{4-\ln3}=9.31\ldots\)

Therefore

\(\frac{dx}{dt}=9.31\) to 3 significant figures.

This completes the solution and gives the required result.