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

how to find the second derivative of quotient rule?

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

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

f"'(x) : ???

- calculus -
**Becky**, Monday, August 13, 2012 at 5:21am
sorry 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 9:08am
the 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 10:52am
The 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 12:22pm
You 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 12:25pm
pls 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 12:31pm
i got the answer!! thank you!!!

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