Answer: \(\dfrac{dy}{dx}=\dfrac{2x^2+1}{(x^2+1)^{1/2}}\); no stationary points because \(\dfrac{dy}{dx}\gt 0\) for all real \(x\).

Write

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

Using the product rule,

\(\frac{dy}{dx}=(x^2+1)^{1/2} +x\cdot\frac12(x^2+1)^{-1/2}\cdot2x.\)

So

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

Put over a common denominator:

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

Hence \(a=2\), \(b=1\) and \(p=\frac12\).

The numerator \(2x^2+1\) is always positive, and the denominator \((x^2+1)^{1/2}\) is also always positive. Therefore

\(\frac{dy}{dx}\gt 0\)

for all real \(x\), so the graph has no stationary points.