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Schrodinger
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This isn't the question itself, but I need to find the derivative of something to do the question (Problem below)
 one year ago
 one year ago
Schrodinger Group Title
This isn't the question itself, but I need to find the derivative of something to do the question (Problem below)
 one year ago
 one year ago

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Schrodinger Group TitleBest ResponseYou've already chosen the best response.1
\[y = x ^{2/3}(2+x).\] What i'm doing is applying the Product Rule for Derivatives like such: \[(x ^{2/3})*\frac{ dy }{ dx }(2+x) + (2+x)\frac{ dy }{ dx }(x ^{2/3})\] From there, I get\[x ^{2/3}+\frac{ 2(2+x) }{ 3\sqrt[3]{x} }\]What i'm doing is then multiplying everything by the denominator and ending up with\[5x+4\] after simplifying, but the answer should be\[\frac{ 5x+4 }{ 3\sqrt[3]{} }\]
 one year ago

Schrodinger Group TitleBest ResponseYou've already chosen the best response.1
Whoops. Denominator should be \[3\sqrt[3]{x}\]
 one year ago

Schrodinger Group TitleBest ResponseYou've already chosen the best response.1
Why am I getting \[5x+4\] as opposed to\[\frac{ 5x+4 }{ 3\sqrt[3]{x} }\] ?
 one year ago

Ahmad_Shadab Group TitleBest ResponseYou've already chosen the best response.0
just take the LCM of denominator and u wl get the answer
 one year ago

Schrodinger Group TitleBest ResponseYou've already chosen the best response.1
That didn't really help, but let me specifically go through multiplying through the denominator so somebody can see if i'm doing something wrong: \[x ^{2/3}(3\sqrt[3]{x}) = x ^{2/3}*x ^{1/3}*3=x ^{1}*3=x*3=3x\]Right? Then, since the numerator has been multiplied out, I just have \[3x+2(x+2)\]which just simplifies to\[5x+4\]?
 one year ago

Ahmad_Shadab Group TitleBest ResponseYou've already chosen the best response.0
u got the correct numerator but u forgot the denominator
 one year ago

Schrodinger Group TitleBest ResponseYou've already chosen the best response.1
What does that mean, dude? I know that. You're kind of unclear.
 one year ago

Schrodinger Group TitleBest ResponseYou've already chosen the best response.1
I'm just curious why, unless I broke math, lol, my method isn't as valid.
 one year ago

stamp Group TitleBest ResponseYou've already chosen the best response.1
\[y=x^{2/3}(2+x)\] u = x^2/3, du = 2/3 x^1/3 v = 2 + x, dv = 1 u dv + v du = y' x^2/3 (1) + (2 + x) 2/3 x^1/3 \[y'=x^{2/3}+\frac{2(2+x)}{3x^{1/3}}\]\[y'=\frac{3x+4+2x}{3x^{1/3}}\]\[y'=\frac{5x+4}{3x^{1/3}}\]verification  http://www.wolframalpha.com/input/?i=derivative+of+x^%282%2F3%29%282%2Bx%29 post a line number if you want to go over any steps more in depth
 one year ago

Schrodinger Group TitleBest ResponseYou've already chosen the best response.1
Line number?
 one year ago

Schrodinger Group TitleBest ResponseYou've already chosen the best response.1
(I dunno what that means.) I checked the wolfram page to see if you were referring to the step number in a stepbystep solution, but I don't see one. PS, thank you
 one year ago

Schrodinger Group TitleBest ResponseYou've already chosen the best response.1
OH. Derp. Just got how your method works. But why is mine still wrong? Shouldn't it just be another way to arrive at the same answer?
 one year ago

stamp Group TitleBest ResponseYou've already chosen the best response.1
yours is ok but you combine the terms, so you need to get a common denominator of 3x ^ 1/3
 one year ago

ZeHanz Group TitleBest ResponseYou've already chosen the best response.0
I would do this:\[y'=\frac{2(2+x)}{3\sqrt[3]{x}}+\sqrt[3]{x^2}=\frac{2(2+x)}{3\sqrt[3]{x}}+\frac{\sqrt[3]{x^2}}{1}\cdot \frac{3\sqrt[3]{x}}{3\sqrt[3]{x}}=\frac{2(2+x)}{3\sqrt[3]{x}}+\frac{3x}{3\sqrt[3]{x}}=\]\[y'=\frac{ 5x +4}{ 3\sqrt[3]{x} }\]
 one year ago

stamp Group TitleBest ResponseYou've already chosen the best response.1
\[x^{2/3}=\frac{3x}{3x^{1/3}}\]
 one year ago

stamp Group TitleBest ResponseYou've already chosen the best response.1
3x + 2(2+x) = 3x + 4 + 2x = 5x + 4 thats you numerator
 one year ago

Schrodinger Group TitleBest ResponseYou've already chosen the best response.1
Give me just a minute to mull over this, mind being blown.
 one year ago

Schrodinger Group TitleBest ResponseYou've already chosen the best response.1
Ugugugugugugugugh. I still don't understand what's wrong with multiplying out the denominator. If I multiply out to the denominator like this: \[3\sqrt[3]{x}[\frac{ x ^{2/3} }{ 1 }+\frac{ 2(2+x) }{ 3\sqrt[3]{x} }]\]Shouldn't the denominator of the second term disappear, and the denominator of the first term stay the same since you're multiplying by \[\frac{ 3\sqrt[3]{x} }{ 1 }\]?
 one year ago

stamp Group TitleBest ResponseYou've already chosen the best response.1
\[x^{2/3}*3x^{1/3}/3x^{1/3}=3x/3x^{1/3}\]
 one year ago

stamp Group TitleBest ResponseYou've already chosen the best response.1
\[3x/3x^{1/3}+2(2+x)/3x^{1/3}=(5x+4)/3x^{1/3}\]
 one year ago

Meepi Group TitleBest ResponseYou've already chosen the best response.0
The problem with multiplying it out like you did \[ y' = x ^{2/3}+\frac{ 2(2+x) }{ 3\sqrt[3]{x} }\] Multiplying by \(\Large \frac{ 3\sqrt[3]{x} }{ 1 }\) gives \[\frac{ 3\sqrt[3]{x} }{ 1 }y' = \frac{ 3\sqrt[3]{x} }{ 1 }\left(x ^{2/3}+\frac{ 2(2+x) }{ 3\sqrt[3]{x} }\right)\] So instead of the y' you were looking for, you calculated \(\Large \frac{ 3\sqrt[3]{x} }{ 1 }y' \)!
 one year ago

Schrodinger Group TitleBest ResponseYou've already chosen the best response.1
I'm confused about why you multiplied by \[\frac{ 3x ^{1/3} }{ 3x ^{1/3} } \]As opposed to\[\frac{ 3x ^{1/3} }{ 1 }\]? If I multiplied by the second term, wouldn't the denominator of the second term cancel out, and since the term i'm multiplying by is over one, it should only affect the numerator of the second term? I feel like I fundamentally learned something wrong in Algebra, and i'm just now figuring it out.
 one year ago

Schrodinger Group TitleBest ResponseYou've already chosen the best response.1
@Meepi: Just saw your reponse, gimme a sec to read, my bad
 one year ago

Schrodinger Group TitleBest ResponseYou've already chosen the best response.1
Okay. World class screwup. I'm pretty sure over literally the history of my doing math i've been constantly assuming functions to be equal to zero even though they're clearly not and it's usually not consequential unless something like this happens. Jesus. Thanks.
 one year ago

Schrodinger Group TitleBest ResponseYou've already chosen the best response.1
*headdesk*
 one year ago

Schrodinger Group TitleBest ResponseYou've already chosen the best response.1
Yup, better just not do that EVER again, lmao.
 one year ago
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