differentiating ln. (1 Viewer)

[atBondi]

diff:
ln [ ( 3x^2 + 1 ) ( 2x + 5) ]

why cant i use the product rule, and then have f'(x) / f(x)? i get a different answer to when i split it up into sum of 2 logs.

 

[PwnerKebab]

You shouldnt get a different answer.

ln [ ( 3x^2 + 1 ) ( 2x + 5) ]

by the product rule:

d/dx { ln [ ( 3x^2 + 1 ) ( 2x + 5) ] }

= 2(3x^2 + 1) + 6x(2x+5) / ( 3x^2 + 1 ) ( 2x + 5)

or alternatively:

ln [ ( 3x^2 + 1 ) ( 2x + 5) ]

= ln ( 3x^2 + 1 ) + ln ( 2x + 5)

d/dx { ln ( 3x^2 + 1 ) + ln ( 2x + 5) }

= 6x/(3x^2 + 1) + 2/(2x+5)

cross multiply and you get the same as using the product rule.

Get Better.

 

[atBondi]

awww, it does work... turns out the solution was wrong...
Oh well, i guess i should of picked that up myself.. thanks anyway!

 

[PwnerKebab]

Oh well, i guess i should of picked that up myself

Yeah, you're not good.

 

[atBondi]

I'm sorry.