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Schrodinger
 3 years ago
This isn't the question itself, but I need to find the derivative of something to do the question (Problem below)
Schrodinger
 3 years ago
This isn't the question itself, but I need to find the derivative of something to do the question (Problem below)

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Schrodinger
 3 years ago
Best 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\[\frac{ 5x+4 }{ 3\sqrt[3]{} }\]

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0just take the LCM of denominator and u wl get the answer

Schrodinger
 3 years ago
Best ResponseYou've already chosen the best response.1That 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\[5x+4\]?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0u got the correct numerator but u forgot the denominator

Schrodinger
 3 years ago
Best ResponseYou've already chosen the best response.1What does that mean, dude? I know that. You're kind of unclear.

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

stamp
 3 years ago
Best 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

Schrodinger
 3 years ago
Best 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

Schrodinger
 3 years ago
Best ResponseYou've already chosen the best response.1OH. 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?

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

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

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

stamp
 3 years ago
Best ResponseYou've already chosen the best response.13x + 2(2+x) = 3x + 4 + 2x = 5x + 4 thats you numerator

Schrodinger
 3 years ago
Best ResponseYou've already chosen the best response.1Give me just a minute to mull over this, mind being blown.

Schrodinger
 3 years ago
Best ResponseYou've already chosen the best response.1Ugugugugugugugugh. 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 \[\frac{ 3\sqrt[3]{x} }{ 1 }\]?

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The 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 \(\Large \frac{ 3\sqrt[3]{x} }{ 1 }y' \)!

Schrodinger
 3 years ago
Best ResponseYou've already chosen the best response.1I'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.

Schrodinger
 3 years ago
Best ResponseYou've already chosen the best response.1@Meepi: Just saw your reponse, gimme a sec to read, my bad

Schrodinger
 3 years ago
Best ResponseYou've already chosen the best response.1Okay. 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.

Schrodinger
 3 years ago
Best ResponseYou've already chosen the best response.1Yup, better just not do that EVER again, lmao.
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