Answer: \(\dfrac{dy}{dx}=\dfrac{(2x^2+10)^{1/2}}{(x-1)^2}(4x^2-6x-10)\), and the only stationary point for \(x\gt 1\) is at \(x=\dfrac52\).

(a) Let

\(y=\frac{(2x^2+10)^{3/2}}{x-1}.\)

Differentiate using the quotient rule. First,

Thus

Factor out \((2x^2+10)^{1/2}\):

Simplify the bracket:

\(6x(x-1)-(2x^2+10)=6x^2-6x-2x^2-10=4x^2-6x-10.\)

Therefore

So \(A=4\), \(B=-6\) and \(C=-10\).

(b) For \(x\gt 1\), the factor

\(\frac{(2x^2+10)^{1/2}}{(x-1)^2}\)

is positive and never zero. Therefore stationary points occur when

\(4x^2-6x-10=0.\)

Divide by \(2\):

\(2x^2-3x-5=0.\)

Factorising,

\((2x-5)(x+1)=0.\)

So

\(x=\frac52\quad\text{or}\quad x=-1.\)

The curve is defined only for \(x\gt 1\), so \(x=-1\) is rejected. Hence there is exactly one stationary point, and it occurs at

\(x=\frac52.\)