Just want to check whether my answer is correct for quotient rule
F(v) : (600^3/2 + 20250)/v
F'(v) : (300^3/2 - 20250)/v^2
F"(v) : (-150v^5/2 + 40500)/v^3
Can anyone let me know whether my answer is correct for first and second derivative based on F(v) ??
Steve already went through this with you in
http://www.jiskha.com/display.cgi?id=1344880951
Look at
F(v) : (600^3/2 + 20250)/v
what we are saying is that we suspect that that term would be 600v^(3/2)
I got (-150v^(3/2) + 40500)/v^3
As I told you in an earlier post, I would have simplified the original to
f(x) = 600 v^(1/2) + 20250/v and I would not have used the quotient rule, but our answer should come out the same .
My way: ....
f(x) = 600 v^(1/2) + 20250 v^-1
f'(x) = (1/2)600v^(-1/2) - 20250v^-2
= 300 v^(-1/2) - 20250v^-2
which is what you would get if we simplified your answer.
then f''(x) = -150v^(-3/2) + 40500v^-3
Your answer should have been
(-150v^(3/2) + 40500)/v^3
Ok but the power 5/2 is wrong? Should have been power 3/2?
I keep on getting the answer 5/2.. I don't know which part of my working is wrong
I do not want the simplified answer by the way...I need the answer as an expression only
based on my question, i retyped...
F(v) : (600v^3/2 + 20250)/v
F'(v) : (300v^3/2 - 20250)/v^2
F"(v) : (-150v^5/2 + 40500)/v^3
Reiny, isn't your answer suppose to be
f(x) = 600 v^(3/2) + 20250 v^-1,
instead of f(x) = 600 v^(1/2) + 20250 v^-1 ??
1/2 power and 3/2 power there's a difference....
hi i think i know what you mean already.
my answer for F"(v)
(-150v^5/2 + 40500v)/v^4
i simplified further, it became
(-150v(v^3/2 - 270))/v^4
cancel off the common factor v
it became (-150(v^3/2 - 270))/v^3
Am i correct??
To verify if your answers for the first and second derivatives using the quotient rule are correct, we'll apply the quotient rule and compare the results with the derivatives you've provided.
Let's start with the first derivative, F'(v).
The quotient rule states:
If F(v) = u(v) / v(v), where u(v) and v(v) are functions of v, then the derivative of F(v), denoted as F'(v), is given by:
F'(v) = (u'(v) * v(v) - u(v) * v'(v)) / [v(v)]^2
Now, let's compare your answer with the application of the quotient rule:
Given F(v) = (600^(3/2) + 20250) / v
u(v) = 600^(3/2) + 20250
u'(v) = (3/2) * 600^(1/2)
v(v) = v
v'(v) = 1
Applying the quotient rule:
F'(v) = [(u'(v) * v(v)) - (u(v) * v'(v))] / [v(v)]^2
= [((3/2) * 600^(1/2) * v) - ((600^(3/2) + 20250) * 1)] / v^2
= (3/2 * 600^(1/2) * v - (600^(3/2) + 20250)) / v^2
Comparing this with your answer: (300^(3/2) - 20250) / v^2
We can see that there is an error in your answer. You forgot to multiply the term 300^(3/2) by v.
Now, let's move on to the second derivative, F"(v).
Using the quotient rule, the second derivative is found by taking the derivative of F'(v) (the first derivative we just obtained) with respect to v.
Taking the derivative of F'(v):
F"(v) = [d/dv((3/2 * 600^(1/2) * v) - (600^(3/2) + 20250))] / v^2
The derivative of (3/2 * 600^(1/2) * v) with respect to v is:
d/dv((3/2 * 600^(1/2) * v) = 3/2 * 600^(1/2)
Now we have:
F"(v) = [3/2 * 600^(1/2)] / v^2
Comparing this with your answer: (-150v^(5/2) + 40500) / v^3
We can see that there is an error in your answer. The term (-150v^(5/2)) should be (-150v^(1/2)) instead.
In summary, your answer for the first derivative (F'(v)) contains a missing term, and your answer for the second derivative (F"(v)) has an incorrect exponent.