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

- Schrodinger

- jamiebookeater

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- Schrodinger

\[y = x ^{2/3}(2+x).\]
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

Whoops. Denominator should be \[3\sqrt[3]{x}\]

- Schrodinger

Why am I getting \[5x+4\]
as opposed to\[\frac{ 5x+4 }{ 3\sqrt[3]{x} }\]
?

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## More answers

- anonymous

just take the LCM of denominator and u wl get the answer

- Schrodinger

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\[5x+4\]?

- anonymous

u got the correct numerator but u forgot the denominator

- Schrodinger

What does that mean, dude? I know that. You're kind of unclear.

- Schrodinger

I'm just curious why, unless I broke math, lol, my method isn't as valid.

- stamp

\[y=x^{2/3}(2+x)\]
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

Line number?

- Schrodinger

(I dunno what that means.) I checked the wolfram page to see if you were referring to the step number in a step-by-step solution, but I don't see one. PS, thank you

- Schrodinger

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?

- stamp

yours is ok but you combine the terms, so you need to get a common denominator of 3x ^ 1/3

- ZeHanz

I would do this:\[y'=\frac{2(2+x)}{3\sqrt[3]{x}}+\sqrt[3]{x^2}=\frac{2(2+x)}{3\sqrt[3]{x}}+\frac{\sqrt[3]{x^2}}{1}\cdot \frac{3\sqrt[3]{x}}{3\sqrt[3]{x}}=\frac{2(2+x)}{3\sqrt[3]{x}}+\frac{3x}{3\sqrt[3]{x}}=\]\[y'=\frac{ 5x +4}{ 3\sqrt[3]{x} }\]

- stamp

\[x^{2/3}=\frac{3x}{3x^{1/3}}\]

- stamp

3x + 2(2+x) = 3x + 4 + 2x = 5x + 4
thats you numerator

- Schrodinger

Give me just a minute to mull over this, mind being blown.

- stamp

lolk

- Schrodinger

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 \[\frac{ 3\sqrt[3]{x} }{ 1 }\]?

- stamp

\[x^{2/3}*3x^{1/3}/3x^{1/3}=3x/3x^{1/3}\]

- stamp

\[3x/3x^{1/3}+2(2+x)/3x^{1/3}=(5x+4)/3x^{1/3}\]

- anonymous

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 \(\Large \frac{ 3\sqrt[3]{x} }{ 1 }y' \)!

- Schrodinger

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.

- Schrodinger

@Meepi: Just saw your reponse, gimme a sec to read, my bad

- Schrodinger

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.

- Schrodinger

*headdesk*

- stamp

it is ok

- Schrodinger

Yup, better just not do that EVER again, lmao.

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