anonymous
  • anonymous
Find dy/dx step by step? please? :)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[y=ln(\frac{ x^4e^{3x} }{ \sqrt{2x^5-1} })\]
anonymous
  • anonymous
yuck!!!
anonymous
  • anonymous
I know ;(

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anonymous
  • anonymous
i think it would be better to do logarithmic differentiation...
anonymous
  • anonymous
\(\large y=ln(\frac{x^4e^{3x}}{(2x-5)^{\frac{1}{2}}}) = 4lnx + 3x - \frac{1}{2}(2x^5-1)\) simplifying y that way should be much simpler...
anonymous
  • anonymous
i guess that's not so hard afterall....
anonymous
  • anonymous
wait wait.. how'd you get 4ln(x)+3x-1/2(2x^5-1)
anonymous
  • anonymous
use the logaritm properties...
anonymous
  • anonymous
I need to look those up ;/ and review them.
anonymous
  • anonymous
\(\large ln(AB)=lnA+lnB \); \(\large ln(A/B)=lnA-lnB \); \(\large ln(A^B)=BlnA \); hmm... i think that's all of 'em...
anonymous
  • anonymous
Awe, well thank you. Okay so now back to what you got.. Let me look over that real quick
anonymous
  • anonymous
yeah... should be easier now....
anonymous
  • anonymous
Okay so I see you used the second one.. But, I'm confused how you got those values... ln(A)=4lnx+3x when A= x^4e^{3x}
anonymous
  • anonymous
is it..ln(x^4)=4lnx
anonymous
  • anonymous
\(\large y=ln(\frac{x^4e^{3x}}{(2x-5)^{\frac{1}{2}}}) =ln(x^4e^{3x})-ln(2x-5)^{1/2} \)
anonymous
  • anonymous
that's the first step...
anonymous
  • anonymous
now.... \(\large ln(x^4e^{3x})=lnx^4+lne^{3x}=4lnx+3x \) do u undestand this part?
anonymous
  • anonymous
ooo... sorry my bad... the second part is wrong....
anonymous
  • anonymous
ohhh... ln(A^b)=b*ln(A) ln(x^4)=4*ln(x) and e&ln cancel out leaving 3x
anonymous
  • anonymous
\(\large ln(2x-5)^{1/2}=\frac{1}{2}ln(2x-5) \) that's what it should have been...
anonymous
  • anonymous
and yes (to your last post)
anonymous
  • anonymous
lemme rewrite what i shouldve written....
anonymous
  • anonymous
\(\large y=ln(\frac{x^4e^{3x}}{(2x-5)^{\frac{1}{2}}}) = 4lnx + 3x - \frac{1}{2}\color {red}{ln}(2x^5-1) \)
anonymous
  • anonymous
\[\frac{ 4 }{ x }+3..\]
anonymous
  • anonymous
yep... keep going...
anonymous
  • anonymous
I need to do chain rule?
anonymous
  • anonymous
yes, chain rule: \(\large [lny]'=\frac{y'}{y} \)
anonymous
  • anonymous
so \(\large [ln(2x^5-1)]'=\frac{[2x^5-1]'}{2x^5-1} \) simplify....
anonymous
  • anonymous
\[\frac{ 4 }{ x }+3-\frac{ 1 }{ 2 }\frac{ 10x^4 }{ 2x^5-1 } = \frac{ 4 }{ x }+3-\frac{ 5x^4 }{ 2x^5-1 }\]
anonymous
  • anonymous
haha... looks good..:)
anonymous
  • anonymous
why haha? did I do something wrong?
anonymous
  • anonymous
no... i was getting the wrong answers until at the last moment when u posted ur answer... it confirmed my last answer...
anonymous
  • anonymous
oh lol :) you just discouraged me! >;[ no im jk ;) thank you!
anonymous
  • anonymous
oh wait...
anonymous
  • anonymous
haha... just kidding.... yw... :)
anonymous
  • anonymous
haha! jerk ;D
anonymous
  • anonymous
:)
anonymous
  • anonymous
i forgot... nice work on your part... way to stick with the problem....
anonymous
  • anonymous
what you mean? I always finish ;D

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