#### To determine

**To find:** The value of S′(100) and interpret the result.

#### Answer

The value of S′(100) is approximately 0.36 square meters.

#### Explanation

**Given:**

The number of tree species *S* in area *A* is S(A)=0.882A0.842, where *A* is measured in square meters.

**Derivative rules:**

Power Rule: ddx(xn)=nxn−1

**Calculation:**

Obtain the derivative of S(A) with respect to *A.*

S′(A)=dSdA=ddA(0.882A0.842)

=0.882ddA(A0.842)

Apply the power rule and simplify the terms,

S′(A)=0.882(0.842A0.842−1)=(0.882)⋅(0.842)⋅A−0.158=0.742644A−0.158

Therefore, the derivative of *S(A)* is S′(A)=0.742644A−0.158.

Obtain the value of S′(100).

Substitute 100 for *A* in S′(A)=0.742644A−0.158,

S′(100)=0.742644(100−0.158)=0.742644×0.48306≈0.35874

Thus, the value of S′(100) is approximately 0.36.

Since the derivative of S(A) is the rate of change of number of tree species in the area *A*, the rate of change of number of tree species when A=100 square meters is approximately 0.36.