Answer:
(a)The rate of change of price to the number of units sold when 40 units are sold is -$5.28.(decrease)
(b)The rate of change of price to the number of units sold when 40 units are sold is -$1.89.(decrease)
(c)The p' is decreasing at 40 units.
Explanation:
Given function is

where p is the price per unit dollar when x units are demanded.

Differentiating with respect to x



Again differentiating with respect to x

![\Rightarrow p''=(4000[(ln(x+10))^2.1+(x+10).(2(ln(x+10)))/((x+10))])/((x+10)^2(ln(x+10))^4)](https://img.qammunity.org/2021/formulas/mathematics/college/pi5xkj5h2186pq7hoqzse9a9o7gqpjbl6s.png)
![\Rightarrow p''=\frac{4000[(ln(x+10))^2+{2(ln(x+10))}]}{(x+10)^2(ln(x+10))^4}](https://img.qammunity.org/2021/formulas/mathematics/college/kp9r6rw4g9sin2lkiso5s1j9zi0clvx64x.png)
![\Rightarrow p''=(4000[(ln(x+10))+2])/((x+10)^2(ln(x+10))^3)](https://img.qammunity.org/2021/formulas/mathematics/college/i3fgnrn8sp52uxel7fcxsup0sbs6gs3e37.png)
(a)
Now,

=- $5.28
The rate of change of price to the number of units sold when 40 units are sold is - $5.28.
(b)

= -$1.89
The rate of change of price to the number of units sold when 40 units are sold is -$1.89.
(c)
![p''|_(x=40)=(4000[(ln(40+10))+2])/((40+10)^2(ln(40+10))^3)>0](https://img.qammunity.org/2021/formulas/mathematics/college/cquiwyg39xq8t2wkl73ycoqcnoa1eck0jn.png)
Since p''>0.
The rate is decreasing at 40 units.