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Posted by **Becky** on Monday, August 13, 2012 at 6:12pm.

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**, Monday, August 13, 2012 at 8:11pmSteve 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**, Monday, August 13, 2012 at 10:38pmOk 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**, Monday, August 13, 2012 at 10:49pmI do not want the simplified answer by the way...I need the answer as an expression only

- calculus -
**Becky**, Tuesday, August 14, 2012 at 1:03ambased 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**, Tuesday, August 14, 2012 at 10:09amhi 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??

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