RemarkIf you don't start exactly the right way, you can get into all kinds of trouble. This is just one of those cases. I think the best way to start is to divide both terms by x^(1/2)
Step OneDivide both terms in the numerator by x^(1/2)
y= 6x^(1/2) + 3x^(5/2 - 1/2)
y =6x^(1/2) + 3x^(4/2)
y = 6x^(1/2) + 3x^2 Now differentiate that. It should be much easier.
Step TwoDifferentiate the y in the last step.
y' = 6(1/2) x^(- 1/2) + 3*2 x^(2 - 1)
y' = 3x^(-1/2) + 6x I wonder if there's anything else you can do to this. If there is, I don't see it.
I suppose this is possible.
y' = 3/x^(1/2) + 6x
y' =
Frankly I like the first answer better, but you have a choice of both.
Firstly, let's take first factor:
(y^3)^2 = y^(3*2) = y^6
and then:
y^6 * y^7 = y^(6 + 7) = y^13
all students = 150
M = 60
S = 45
M and S = 25
(a) At least one of the two requirements:
M or S = M + S - (M and S) = 60 + 45 - 25 = 80
(b) Exactly one of the two requirements:
(M or S) - (M and S) = 80 - 25 = 55
(c) Neither requirement:
(all students) - (M or S) = 150 - 80 = 70
Standard form
im pretty sure
Answer:
7z-1 hope this helps
Step-by-step explanation: