Asked by Becky

how to find the second derivative of quotient rule?
f(x): 700v^2 + 3450/v
f'(x) : 700v^2 -3450/v^2
f"'(x) : ???

Answers

Answered by Becky
sorry typo error above, question is,

f(x): 700v^2 + 34500/v
f'(x) : 700v^2 -34500/v^2
f"'(x) : ???
Answered by Reiny
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
Answered by Becky
The question should be:


f(x): 700v^2 + 34500/v
f'(x) : 700v^2 -34500/v^2
f''(x) : ???
Answered by Reiny
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
Answered by Becky
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) : ???

Answered by Becky
i got the answer!! thank you!!!
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