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anonymous
 4 years ago
Find dy/dx
step by step? please? :)
anonymous
 4 years ago
Find dy/dx step by step? please? :)

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[y=ln(\frac{ x^4e^{3x} }{ \sqrt{2x^51} })\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i think it would be better to do logarithmic differentiation...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\(\large y=ln(\frac{x^4e^{3x}}{(2x5)^{\frac{1}{2}}}) = 4lnx + 3x  \frac{1}{2}(2x^51)\) simplifying y that way should be much simpler...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i guess that's not so hard afterall....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait wait.. how'd you get 4ln(x)+3x1/2(2x^51)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0use the logaritm properties...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I need to look those up ;/ and review them.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\(\large ln(AB)=lnA+lnB \); \(\large ln(A/B)=lnAlnB \); \(\large ln(A^B)=BlnA \); hmm... i think that's all of 'em...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Awe, well thank you. Okay so now back to what you got.. Let me look over that real quick

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah... should be easier now....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay 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
 4 years ago
Best ResponseYou've already chosen the best response.0\(\large y=ln(\frac{x^4e^{3x}}{(2x5)^{\frac{1}{2}}}) =ln(x^4e^{3x})ln(2x5)^{1/2} \)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that's the first step...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now.... \(\large ln(x^4e^{3x})=lnx^4+lne^{3x}=4lnx+3x \) do u undestand this part?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ooo... sorry my bad... the second part is wrong....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohhh... ln(A^b)=b*ln(A) ln(x^4)=4*ln(x) and e&ln cancel out leaving 3x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\(\large ln(2x5)^{1/2}=\frac{1}{2}ln(2x5) \) that's what it should have been...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and yes (to your last post)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lemme rewrite what i shouldve written....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\(\large y=ln(\frac{x^4e^{3x}}{(2x5)^{\frac{1}{2}}}) = 4lnx + 3x  \frac{1}{2}\color {red}{ln}(2x^51) \)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ 4 }{ x }+3..\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I need to do chain rule?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes, chain rule: \(\large [lny]'=\frac{y'}{y} \)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so \(\large [ln(2x^51)]'=\frac{[2x^51]'}{2x^51} \) simplify....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ 4 }{ x }+3\frac{ 1 }{ 2 }\frac{ 10x^4 }{ 2x^51 } = \frac{ 4 }{ x }+3\frac{ 5x^4 }{ 2x^51 }\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0haha... looks good..:)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0why haha? did I do something wrong?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no... i was getting the wrong answers until at the last moment when u posted ur answer... it confirmed my last answer...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh lol :) you just discouraged me! >;[ no im jk ;) thank you!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0haha... just kidding.... yw... :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i forgot... nice work on your part... way to stick with the problem....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what you mean? I always finish ;D
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