mriley528 said:
For this problem i used the quotient to come up with a derivative for t, and i came up with t'(5)= 19.444 and t'(6)= 18.210, and the change in herd to -1.2342
In your original post, the function name is "N(t)". Now you are referring to the derivative of some function "t". Are they related, or is this a different (new) exercise?
Looking back at your original post, it appears that you have edited the function. (FYI: Replying is generally better, as people don't "see" edits, nor are they notified of such.) But I'm still not clear on what the function is. As you have written it, the function is as follows:
. . . . .\(\displaystyle \L N(t)\,=\,\frac{10(5\,+\,3t)}{1}\,+\,0.04t\)
But that doesn't make much sense...? I think you mean this instead:
. . . . .\(\displaystyle \L N(t)\,=\,\frac{10(5\,+\,3t)}{1\,+\,0.04t}\)
Please confirm or correct.
When you reply, please show the steps you have tried thus far (not just terminal values, but the in-between working leading up to them). Thank you.
Eliz.