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Find dy/dx step by step? please? :)

Mathematics
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\[y=ln(\frac{ x^4e^{3x} }{ \sqrt{2x^5-1} })\]
yuck!!!
I know ;(

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Other answers:

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

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