Question
how to find the second derivative of quotient rule?
f(x): 700v^2 + 3450/v
f'(x) : 700v^2 -3450/v^2
f"'(x) : ???
f(x): 700v^2 + 3450/v
f'(x) : 700v^2 -3450/v^2
f"'(x) : ???
Answers
Becky
sorry typo error above, question is,
f(x): 700v^2 + 34500/v
f'(x) : 700v^2 -34500/v^2
f"'(x) : ???
f(x): 700v^2 + 34500/v
f'(x) : 700v^2 -34500/v^2
f"'(x) : ???
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
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
Becky
The question should be:
f(x): 700v^2 + 34500/v
f'(x) : 700v^2 -34500/v^2
f''(x) : ???
f(x): 700v^2 + 34500/v
f'(x) : 700v^2 -34500/v^2
f''(x) : ???
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
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
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) : ???
to make it clearer for the question, it is...
f(x): (700v^2 + 34500)/v
f'(x) : (700v^2 -34500)/v^2
f''(x) : ???
Becky
i got the answer!! thank you!!!