calculus
posted by Becky .
Just want to check whether my answer is correct for quotient rule
F(x) : (600^3/2 + 20250)/v
F'(x) : (300^3/2  20250)/v^2
F"(x) : (150v^5/2 + 40500)/v^3
Can anyone let me know whether my answer is correct for first and second derivative based on F(x) ??
Thank you!!

calculus 
Steve
I suspect a typo, since all that junk in parens is just a number, N
F(v) = N/v, or Nv^1
so,
F'(v) = (1)Nv^2 = N/v^2
F''(v) = (2)(N)v^3 = 2Nv^3
I have no idea where that v^5 came from 
calculus  oops 
Steve
Oops. That's 2N/v^3

calculus 
Becky
nope, it's not a typo. the 5/2 is the power of 5 over 2. it is not 5 divide by 2.
is my above answer correct? 
calculus 
Steve
Still doesn't explain where the v^5/2 came from. I don't see any v terms anywhere above there. hat is F(v) really?
If it involves v^3/2, then unless you get rid of the parentheses, you will need to use the quotient rule. 
calculus 
Becky
yes it should be,
F(v) : (600^3/2 + 20250)/v
F'(v) : (300^3/2  20250)/v^2
F"(v) : (150v^5/2 + 40500)/v^3
for F"(v),
this is how i get:
F'(v) : (300^3/2  20250)/v^2
F"(v) : ((300v^3/2  20250)' * v^2)  ((300v^3/2  20250) * (v^2)')/v^4
F"(v) : ((450v^1/2)*v^2)  ((300v^3/2  20250) * 2v)/v^4
F"(v) : (450v^5/2  600v^5/2 + 40500v)/v^4
F"(v) : (150v^5/2 + 40500v)/v^4
F"(v) : (150v^5/2 + 40500)/v^3
which part am i wrong?
is my first derivative correct by the way based on F(v) ?
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