If $y'(t)=.2y(t)(1000-y(t))$, what are the values of $y$ such that $y(t)$ is increasing at a decreasing rate? A population rate changes at a rate modeled by the differential equation 
$ \frac{dy}{dt} = .2y(1000-y),   $ where t is measured in years. What are all y values for which y is increasing at a decreasing rate? 

An answer online claims that the second derivative of this would be:

$ 200 - .4y  $, however, isn't this answer leaving out dy/dt? In other words, should not the second derivative be:

$200 - .4y\frac{dy}{dt}   $?

On this site, the solution does not have the last dy/dt: http://www.rtmsd.org/cms/lib9/PA01000204/Centricity/Domain/211/2013%20BC%20Multiple%20Choice%20SOLUTIONS.pdf

Isn't that incorrect?
We have $\frac{dy}{dt} = .2y(1000-y)$

$\frac{dy}{dt} = 200y - .2y^2$

1.) If finding derivative with respect to t, then it is -

$\frac{d^2y}{dt^2} = 200 \frac{dy}{dt} - 0.4y\frac{dy}{dt}$

2.) If finding derivative with respect to y, then it is -

$= 200 - 0.4y$
Yes it seems that the person who posted that answer took the derivative with respect to y and did not use the chain rule to get the derivative with respect to t.