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!!
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
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?
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.
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) ?
To verify if your answers for the first and second derivative using the quotient rule are correct, we need to differentiate the function step by step and compare the results.
Let's start with the given function:
F(x) = (600^(3/2) + 20250) / v
To find the first derivative (F'(x)), we apply the quotient rule, which states that for two functions f(x) and g(x), their quotient's derivative is calculated as:
(f'(x) * g(x) - f(x) * g'(x)) / (g(x))^2
In this case, f(x) = 600^(3/2) + 20250 and g(x) = v.
1. f'(x) = (3/2)*600^(1/2)*600^1 + 0 = (3/2)*600^(1/2)*600
2. g'(x) = 0, since g(x) = v, a constant with respect to x.
Now, applying the quotient rule:
F'(x) = (f'(x) * g(x) - f(x) * g'(x)) / (g(x))^2
= ((3/2)*600^(1/2)*600 * v - (600^(3/2) + 20250) * 0) / v^2
= (900 * 600^(1/2) * v) / v^2
= (900 * (600^(1/2) / v)
So, your answer for the first derivative F'(x) should be:
F'(x) = (900 * (600^(1/2) / v)
Now, let's move on to finding the second derivative (F''(x)).
To do this, we need to differentiate F'(x) with respect to x.
1. F'(x) = (900 * (600^(1/2) / v)
To find F''(x), we differentiate F'(x) using the quotient rule again:
2. F''(x) = [(f''(x) * g(x) - f'(x) * g'(x)) * (g(x))^2 - (f'(x) * g(x) - f(x) * g'(x)) * 2g(x) * g'(x)] / (g(x))^4
In this case, f'(x) = 900 * (600^(1/2) / v) and g(x) = v.
3. f''(x) = 0, since f'(x) is a constant with respect to x.
Now, applying the quotient rule:
F''(x) = [(f''(x) * g(x) - f'(x) * g'(x)) * (g(x))^2 - (f'(x) * g(x) - f(x) * g'(x)) * 2g(x) * g'(x)] / (g(x))^4
= [0 * v - (900 * (600^(1/2) / v)) * 0 - (0 - (900 * (600^(1/2) / v)) * 2 * v * 0] / v^4
= [0 - 0 - 0] / v^4
= 0
Therefore, your answer for the second derivative F''(x) is indeed correct:
F''(x) = 0
So, to summarize:
F(x) = (600^(3/2) + 20250) / v
F'(x) = (900 * (600^(1/2) / v)
F''(x) = 0
Your answers for the first and second derivative are correct based on the given function F(x) and applying the quotient rule.