Answer: \(\frac{\mathrm dy}{\mathrm dx}=\frac{\frac{4x}{2x^2+1}(x+2)-\ln(2x^2+1)}{(x+2)^2}\), approximate change \(=0.0849h\), and \(\frac{\mathrm dx}{\mathrm dt}=-4.71\).

Use differentiation to find the gradient information needed. Stationary points occur when the derivative is zero, and tangents or normals are found from the gradient at the given point.

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

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

Therefore

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

At \(x=2\),

\(\frac{\mathrm dy}{\mathrm dx}=\frac{\frac{8}{9}(4)-\ln9}{16}=0.08485\ldots.\)

So, for a small increase \(h\) in \(x\), the approximate change in \(y\) is

\(0.0849h.\)

Given that \(y\) is decreasing by \(0.4\) units per second,

\(\frac{\mathrm dy}{\mathrm dt}=-0.4.\)

Using

\(\frac{\mathrm dy}{\mathrm dt}=\frac{\mathrm dy}{\mathrm dx}\frac{\mathrm dx}{\mathrm dt},\)

we get

\(-0.4=0.08485\ldots\frac{\mathrm dx}{\mathrm dt}.\)

Hence

\(\frac{\mathrm dx}{\mathrm dt}=-4.71.\)

This gives the required answer: \(\frac{\mathrm dy}{\mathrm dx}=\frac{\frac{4x}{2x^2+1}(x+2)-\ln(2x^2+1)}{(x+2)^2}\), approximate change \(=0.0849h\), and \(\frac{\mathrm dx}{\mathrm dt}=-4.71\).