Answer
$$S'(100)=\frac{0.743}{100^{0.158}}\approx0.359 (tree/m^2)$$
$S'(100)$ denotes the instantaneous rate of increase of the number of trees every $m^2$ at an area of $100m^2$. At an area of $100m^2$, the instantaneuous rate of increase of the number of trees every $m^2$ is $0.359tree/m^2$.
Work Step by Step
$$S(A)=0.882A^{0.842}$$
1) Find the derivative of $S(A)$ $$S'(A)=\frac{d}{dA}(0.882A^{0.842})$$
$$S'(A)=0.882\times(0.842A^{-0.158})$$
$$S'(A)\approx\frac{0.743}{A^{0.158}}$$ ($S'(A)$ is in $tree/m^2$)
Therefore, $$S'(100)=\frac{0.743}{100^{0.158}}\approx0.359 (tree/m^2)$$
2) The derivative of a function can be seen as the rate of change of that function.
We know that $S(A)$ is the function of the number of tree species S in a given area A.
Therefore, $S'(A)$ would be the rate of change of the number of trees species S every $m^2$ at an area A.
That means $S'(100)$ denotes the instantaneous rate of increase of the number of trees every $m^2$ at an area of $100m^2$. At an area of $100m^2$, the instantaneuous rate of increase of the number of trees every $m^2$ is $0.359tree/m^2$.
