Answer:
Step-by-step explanation:
Hello, first, let's use the product rule.
Derivative of uv is u'v + u v', so it gives:

Now, we distribute the expression of f(x) and find the derivative afterwards.

Hope this helps.
Do not hesitate if you need further explanation.
Thank you
<em>If the rate is a constant, the relationship is proportional.</em> (The rate is the constant of porportionality.)
_____
For two or more points on the function curve, divide the dependent variable value by the independent one. If you get the same result in every case, the relationship is porportional.
___
<em>Note</em>
For some non-proportional relationships, it is possible to find points on the graph that will pass the above test. If you suspect the relationship is actually not one of proportionality, try more points. Check also to make sure that (0, 0) is on the curve.
Hello from MrBillDoesMath!
Answer:
Binomial with a degree of 6 (the second Choice)
Discussion:
(a^3b+9a^2b^2-4ab^5) - (a^3b-3a^2b^2+ab^5) =
(-4ab^5- ab^5) + ( a^3b-a^3b) + ( 9a^2b^2 + 3a^2b^2) =
(-4ab^5- ab^5) + 0 + ( 9a^2b^2 + 3a^2b^2) =
12 a^2 b^2 - 5 a b^5
This is a binomial with degree 6 (degree of last term = 1 + 5 = 6).
Thank you,
MrB