I 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
Just want to check whether my answer is correct for quotient rule
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!!
5 answers
Oops. That's 2N/v^3
nope, 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?
is my above answer correct?
Still 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.
If it involves v^3/2, then unless you get rid of the parentheses, you will need to use the quotient rule.
yes 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) ?
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) ?