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September 30, 2014

September 30, 2014

Posted by **Becky** on Monday, August 13, 2012 at 5:17am.

f(x): 700v^2 + 3450/v

f'(x) : 700v^2 -3450/v^2

f"'(x) : ???

- calculus -
**Becky**, Monday, August 13, 2012 at 5:21amsorry typo error above, question is,

f(x): 700v^2 + 34500/v

f'(x) : 700v^2 -34500/v^2

f"'(x) : ???

- calculus -
**Reiny**, Monday, August 13, 2012 at 9:08amthe way you typed it ....

f'(x) = 1400v - 3450/v^2 or 1400v - 3450v^-2

f''(x) =1400 + 6900v^-3

IF you meant f(x) = (700v^2 + 3450)/v

I would change it to

f(x) = 700v + 3400v^-1

f'(x) = 700 - 3400v^-2

f''(x) = 6800v^-3 or 6800/v^2

- calculus -
**Becky**, Monday, August 13, 2012 at 10:52amThe question should be:

f(x): 700v^2 + 34500/v

f'(x) : 700v^2 -34500/v^2

f''(x) : ???

- calculus -
**Reiny**, Monday, August 13, 2012 at 12:22pmYou missed my point, the issue was not whether it was 3450 or 34500, the point was the use of brackets.

Did you not look at my reply?

simply change 3450 to 34500 and follow the steps.

You derivative would be correct if you had placed it in brackets , such as

f'(x) = (700v^2 - 34500)/v^2

which would reduce to

f'(x) = 700 - 34500/v^2 = 700 - 34500v^-2

then f''(x) = 69000/v^3

- calculus -
**Becky**, Monday, August 13, 2012 at 12:25pmpls ignore the above question...

to make it clearer for the question, it is...

f(x): (700v^2 + 34500)/v

f'(x) : (700v^2 -34500)/v^2

f''(x) : ???

- calculus -
**Becky**, Monday, August 13, 2012 at 12:31pmi got the answer!! thank you!!!

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