Answer: \(\frac{dy}{dx}=\frac{-x^2+2x+15}{3(x+1)^2(x^2-5)^{2/3}}\); stationary \(x=5\).
Let
\(u=(x^2-5)^{\frac13}.\)
Then
\(\frac{du}{dx}=\frac13(x^2-5)^{-\frac23}\cdot2x =\frac{2x}{3}(x^2-5)^{-\frac23}.\)
Since
\(y=\frac{u}{x+1},\)
the quotient rule gives
\(\frac{dy}{dx}=\frac{(x+1)\frac{du}{dx}-u}{(x+1)^2}.\)
Substituting \(u\) and \(\frac{du}{dx}\),
\(\frac{dy}{dx} =\frac{(x+1)\left(\frac{2x}{3}(x^2-5)^{-\frac23}\right)-(x^2-5)^{\frac13}}{(x+1)^2}.\)
Use a common denominator \(3(x+1)^2(x^2-5)^{2/3}\). The numerator becomes
\(2x(x+1)-3(x^2-5).\)
Simplifying,
\(2x(x+1)-3(x^2-5) =2x^2+2x-3x^2+15 =-x^2+2x+15.\)
Therefore
\(\frac{dy}{dx}=\frac{-x^2+2x+15}{3(x+1)^2(x^2-5)^{\frac23}}.\)
So \(A=-1\), \(B=2\) and \(C=15\).
For a stationary point, set the numerator equal to zero:
\(-x^2+2x+15=0.\)
Equivalently,
\(x^2-2x-15=0.\)
Factorising,
\((x-5)(x+3)=0.\)
Since \(x\gt -1\), reject \(x=-3\). Hence the stationary point has
\(x=5.\)
To determine the nature of the stationary point, one valid method is to check the sign of \(\frac{dy}{dx}\) on either side of \(x=5\). If the gradient changes from positive to negative, the point is a maximum; if it changes from negative to positive, the point is a minimum.
Alternatively, find \(\frac{d^2y}{dx^2}\) and substitute \(x=5\). A positive value gives a minimum and a negative value gives a maximum.
