Question

differentiate the function g(r)=r^2(ln(2r+1))

Answer 1

\[g \left( r \right)=r^2\left( \ln \left( 2r+1 \right) \right)\]like this?

Answer 2

you there?

Answer 3

yeah

Answer 4

thats right

Answer 5

so use the product rule

Answer 6

could you break it down for me? Having trouble breaking it apart

Answer 7

I need to know how to solve the problem

Answer 8

u = r^2
v = ln(2r+1)

f(r)=u*v => f'(r) = u'*v + u*v'

Answer 9

what is u'?

Answer 10

2r

Answer 11

and what is v'?

Answer 12

ill take a guess at ln(2+0)?

Answer 13

nope...

\[\frac{ d }{ dr }\ln \left( f \left( r \right) \right)=\frac{ f'\left( r \right) }{ f \left( r \right) }\]

Answer 14

ln(2)/ln(2r+1)?

Answer 15

or just 2/2r+1?

Answer 16

there you go!

Answer 17

2r+1

Answer 18

yeah, the second one is correct...

Answer 19

\[v'=\frac{ 2 }{ 2r+1 }\]

Answer 20

so v' is 1/r+1?

Answer 21

ah ok

Answer 22

now put it all together

Answer 23

2r*ln(2r+1)+r^2*(2)/(2r+1)

Answer 24

looks good!

Answer 25

\[g'\left( r \right)=2r \cdot \ln \left( 2r+1 \right)+\frac{ 2r^2 }{ 2r+1 }\]

Answer 26

excellent! thank you
is this the best /simpliest form?

Answer 27

i guess you could factor...\[g'\left( r \right)=2r \left[ \ln \left( 2r+1 \right) +\frac{ r }{ 2r+1 }\right]\]

Answer 28

thanks!