Question

P=(V^2R)/(R+r)^2
V, R are constants; r is a variable
Find dP/dr

Answer 1

P' = V^2R * d(1/(R+r)^2)/dr
P' = V^2R * d((R+r)^(-2))/dr
P' = V^2R * ( -2*(R+r)^(-3))

Answer 2

\[P=(V^2R).(R+r)^{-2}\]

P' = (V^2 R) -2(R+r)^-3

is what I get

Answer 3

[V^2 R]' = 0

[(R+r)^-2]' = -2(R+r)^-3

Answer 4

According to the software, -2(R+r)^-3 is not the correct answer.

Answer 5

thats becasue it has a negative exponent. I just assumed you had some background in these things

Answer 6

\[P' = -\frac{(V^2 R)}{2(R+r)^3}\]

Answer 7

The software automatically converts the negative exponents, so there was no need for me to do anything with it. Your most recent answer is still coming up wrong.

Answer 8

thats becasue you stated that V was a constant as well

Answer 9

V get derived with respecr to 'r' then as well since it aint a constant

Answer 10

V is a constant. That's what the professor told us in class.

Answer 11

your screenshot does not say V is a constant; have you been using V as a constant in this? or are you just assuming that some problem that your prof used where V was a constant applies to all Vs that you come across?

Answer 12

He told us in class that, for this specific problem, V is a constant. He said the authors of the book failed to include that piece of information in the problem.

Answer 13

Then if you can find where I have an error at, then let me know. Cuase I cant see it :)

Answer 14

Me neither. This problem is driving me nuts. :P

Answer 15

i think I do; put the 2 up top; I dragged it along and it should stay up top

Answer 16

\[-2(R+r)^{-3}=\frac{-2}{(R+r)^3}\]

Answer 17

Yep. That was it. :D

Answer 18

But it did make me keep the \[V ^{2}R\]

Answer 19

lol ..... accursed programs :)

Answer 20

Indeed. Thank you for all your help though!