^{2}. In this case the second term is zero so the answer will be y' = (dx/dx)/k

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

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The derivative of a fraction is: y=x/k => y'= (k(dx/dx) - x(dk/dx))/k^{2}. In this case the second term is zero so the answer will be y' = (dx/dx)/k

P=2RT/3b

(dP/dT)= (2T)/3b

I do not see anything to any power.

Thanks

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.

P=2RT/3b

let u=2RT let v = 3b

du/dx=2R** dv/dx=3**

let u=2RT let v = 3b

du/dx=2R

So, because b is a constant and (3 x b) is a constant, then in this case dv/dx = 0?

Is that reasoning correct?

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...

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

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## Comments

dF(x) = x^b dx

= x^(b-1) / b