Answer: (a) \(A=-4,\ B=2\); (b) approximate increase \(\displaystyle -\frac{2p}{\sqrt[3]{10}}\).

Answer: (a) \(A=-4,\ B=2\); (b) approximate increase \(\displaystyle -\frac{2p}{\sqrt[3]{10}}\).

(a) Write

\(y=\frac{(3x^2-2)^{2/3}}{x-1}.\)

Using the quotient rule,

Factor out \((3x^2-2)^{-1/3}\):

\(\frac{dy}{dx} =\frac{(3x^2-2)^{-1/3}}{(x-1)^2} \left(4x(x-1)-(3x^2-2)\right).\)

Simplify the bracket:

\(4x(x-1)-(3x^2-2)=4x^2-4x-3x^2+2=x^2-4x+2.\)

Therefore

\(\frac{dy}{dx} =\frac{(3x^2-2)^{-1/3}}{(x-1)^2}(x^2-4x+2).\)

So

\(A=-4,\qquad B=2.\)

(b) For a small increase \(p\), the approximate increase in \(y\) is

\(\frac{dy}{dx}\bigg|_{x=2}p.\)

At \(x=2\),

\(3x^2-2=10,\qquad x^2-4x+2=4-8+2=-2,\)

and \((x-1)^2=1\). Hence

\(\frac{dy}{dx}\bigg|_{x=2}=-2(10)^{-1/3}.\)

Therefore the approximate increase is

\(-\frac{2p}{\sqrt[3]{10}}.\)