Answer: The approximate change in \(y\) is \(14.3h\).

Work through the problem step by step, using exact values where possible before giving any final numerical approximation.

Let

\(y=\frac{4x^3+2\sin 8x}{1-x}.\)

Using the quotient rule,

\(\frac{dy}{dx} =\frac{(1-x)(12x^2+16\cos 8x)-(4x^3+2\sin 8x)(-1)}{(1-x)^2}.\)

Hence

\(\frac{dy}{dx} =\frac{(1-x)(12x^2+16\cos 8x)+4x^3+2\sin 8x}{(1-x)^2}.\)

At \(x=0.1\),

\(\frac{dy}{dx} =\frac{0.9(12(0.1)^2+16\cos0.8)+4(0.1)^3+2\sin0.8}{0.9^2}.\)

This gives

\(\frac{dy}{dx}\approx14.2954.\)

For a small increase \(h\) in \(x\),

\(\Delta y\approx \frac{dy}{dx}h.\)

Therefore the approximate change in \(y\) is

\(14.3h.\)