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September 14, 2014

September 14, 2014

Posted by **Becky** on Monday, August 13, 2012 at 2:02pm.

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**, Monday, August 13, 2012 at 3:28pmI 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**, Monday, August 13, 2012 at 3:29pmOops. That's 2N/v^3

- calculus -
**Becky**, Monday, August 13, 2012 at 3:43pmnope, 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**, Monday, August 13, 2012 at 3:57pmStill 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**, Monday, August 13, 2012 at 4:12pmyes 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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