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

calculus 
Reiny,
Steve already went through this with you in
http://www.jiskha.com/display.cgi?id=1344880951
Look at
F(v) : (600^3/2 + 20250)/v
what we are saying is that we suspect that that term would be 600v^(3/2)
I got (150v^(3/2) + 40500)/v^3
As I told you in an earlier post, I would have simplified the original to
f(x) = 600 v^(1/2) + 20250/v and I would not have used the quotient rule, but our answer should come out the same .
My way: ....
f(x) = 600 v^(1/2) + 20250 v^1
f'(x) = (1/2)600v^(1/2)  20250v^2
= 300 v^(1/2)  20250v^2
which is what you would get if we simplified your answer.
then f''(x) = 150v^(3/2) + 40500v^3
Your answer should have been
(150v^(3/2) + 40500)/v^3 
calculus 
Becky,
Ok but the power 5/2 is wrong? Should have been power 3/2?
I keep on getting the answer 5/2.. I don't know which part of my working is wrong 
calculus 
Becky,
I do not want the simplified answer by the way...I need the answer as an expression only

calculus 
Becky,
based on my question, i retyped...
F(v) : (600v^3/2 + 20250)/v
F'(v) : (300v^3/2  20250)/v^2
F"(v) : (150v^5/2 + 40500)/v^3
Reiny, isn't your answer suppose to be
f(x) = 600 v^(3/2) + 20250 v^1,
instead of f(x) = 600 v^(1/2) + 20250 v^1 ??
1/2 power and 3/2 power there's a difference.... 
calculus 
Becky,
hi i think i know what you mean already.
my answer for F"(v)
(150v^5/2 + 40500v)/v^4
i simplified further, it became
(150v(v^3/2  270))/v^4
cancel off the common factor v
it became (150(v^3/2  270))/v^3
Am i correct??