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P=2RT/3b what is dp/dT? (differentiation)

P=2RT/3b what is dp/dT? (differentiation) I know the answer is 2R/3b but can't see the steps to get there


Thnx for answering
I don't get this bit:
if F(x) = x^b, where b = exponent then...
dF(x) = x^b dx
= x^(b-1) / b
I thought if dF(x) of x^b 
= bx^b-1
Sorry, I seem to be getting a lot of these exercises right but this one fails me miserably??
You are absolutely correct. dF(x)  of x^b = bx^b-1
My apologies, it looks like I typed it in wrong.
My (obviously incorrect) working so far:
let u=2RT  let v = 3b
du/dx=2R     dv/dx=3
Quotient rule
(v*du/dx - udv/dx) / v^2
(3b*2R) - (2RT*3b) / 9b^2
Hi all,
Just wanted to thank all those who answered, was a great help to me and is really appreciated.
Beers all round.  :o)

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Francisco E. | Francisco; Civil Engineering, Math., Science, Spanish, Computers.Francisco; Civil Engineering, Math., Sci...
5.0 5.0 (1 lesson ratings) (1)
The derivative of a fraction is: y=x/k => y'= (k(dx/dx) - x(dk/dx))/k2. In this case the second term is zero so the answer will be y' = (dx/dx)/k
(dP/dT)= (2T)/3b
I do not see anything to any power.



I don't see why the second term is zero though. I have included what I though was right in a comment to my original question.

I am not seeing the wood for the trees here I fear...

The second term of the derivative is Zero because b in this case is a constant and the derivative of a constant is equal tozero. I hope I explained and answered your question.
Ah, I think I may have seen a glimmer.
let u=2RT  let v = 3b
du/dx=2R     dv/dx=3
So, because b is a constant and (3 x b) is a constant, then in this case dv/dx = 0?
Is that reasoning correct?
Alec T. | Physics Experimentalist Specializing in Math and Science TutoringPhysics Experimentalist Specializing in ...
5.0 5.0 (11 lesson ratings) (11)
Rearrange P=2RT/3b into the form
P = 2R/3b * T
This looks an awful lot like y = mx! If y = mx, then we know the curve is a line, and the derivative is the slope, dy/dx = m.
In your problem, P is y, T is x, and m is 2R/3b.
so dP/dT = 2R/3b!
This is simply done this way because R and B are not also functions of T.
Ethan G. | Engineering Grad for Math and Science TutoringEngineering Grad for Math and Science Tu...
So, another way to think about this problem is "What is the change of P with respect to change of T?" Since the only two variables we are looking at is P and T, R and b are constants, thus don't change, thus they can be "ignored".

So when you differentiate P with respect to T, remember

if F(x) = x^b, where b = exponent then...
 dF(x) = x^b dx
         = bx^(b-1)
So in this case, dP/dT ==>
dP = T dT  where b = 1
    = (1)T^(1-1)
    = T^(0)
    = 1
So dP/dT = 2R(1) / 3b