# calculus

posted by on .

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!!

• calculus - ,

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

• calculus -- oops - ,

Oops. That's 2N/v^3

• calculus - ,

nope, it's not a typo. the 5/2 is the power of 5 over 2. it is not 5 divide by 2.

• calculus - ,

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.

• calculus - ,

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) ?