I am trying to linearize this system, but I am having difficulty matching my result to the stated result because of the complexity of the terms.
The system is given by:
$f(x,u) = \begin{bmatrix} \dfrac{x_{2}}{x_{2} + K}x_{1} - Dx_{1}\\ -\dfrac{x_{2}}{x_{2} + K}x_{1} + D(1-x_{2}) \end{bmatrix}$
There are two equilibria, I am interested in linearizing about $x_{e} = [0 \ 1]^{T}$
The stated solution is:
$\frac{\partial{f(x,u)}}{\partial{x}} = $ $\begin{bmatrix} \dfrac{x_{2e}}{x_{2e} + K} - D & \dfrac{Kx_{1e}}{(x_{2e} + K)^{2}} \\ -\dfrac{x_{2e}}{x_{2e} + K} & -\dfrac{Kx_{1e}}{x_{2e} + K} - D \end{bmatrix} |x_{e}$
With substitution of $x_{e}$ and K = $\dfrac{1-D}{D}$ leading to:
$\begin{bmatrix} 0 & 0 \\ -D & - D \end{bmatrix}$
I am skeptical about the result, to me it seems the linearization requires to use of quotient/product rule which generates extra terms that do not drop out.
The Jacobian is given as:
$$ \boldsymbol{J}=\left[ \begin {array}{cc} {\frac {x_{2}}{x_{2}+K}}-D&{\frac {x_{1}}{x _{2}+K}}-{\frac {x_{2}\,x_{1}}{ \left( x_{2}+K \right) ^{2}}} \\-{\frac {x_{2}}{x_{2}+K}}&-{\frac {x_{1}}{x_{2}+K }}+{\frac {x_{2}\,x_{1}}{ \left( x_{2}+K \right) ^{2}}}-D\end {array} \right] $$
Evaluated at $x_1=0$ and $x_2=1$ you will obtain:
$$\boldsymbol{A}_{\text{eq}}(K,D)=\left[ \begin {array}{cc} \left( 1+K \right) ^{-1}-D&0 \\ - \left( 1+K \right) ^{-1}&-D\end {array} \right].$$
Using $K=(1-D)/D$: $$\boldsymbol{A}_{\text{eq}}(D)=\left[ \begin {array}{cc} 0&0\\ -D&-D\end {array} \right] $$