# calculus

posted by Becky

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) ??

1. 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

then f''(x) = -150v^(-3/2) + 40500v^-3

(-150v^(3/2) + 40500)/v^3

2. 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

3. Becky

I do not want the simplified answer by the way...I need the answer as an expression only

4. 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

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....

5. Becky

hi i think i know what you mean already.
(-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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