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Carissa15
 one year ago
how do I differentiate 4sqrt(u^2+u) ?
Carissa15
 one year ago
how do I differentiate 4sqrt(u^2+u) ?

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arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.1with respect to u?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0WELCOME TO OPENSTUDY!

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0Oh yes, sorry. It says Differentiate f(u) I have not yet differentiated using roots yet

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so first of all do you know how to simplify the equation?

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.1See this......

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.1@Carissa15 you there?

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0yes, that makes sense so far

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0I thought that you add 1/2 to "remove" the root from the equation but then I get lost..

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0So you could then take the derivative of the simplified equation? I didn't know how to simplify and remove the \[\sqrt{u}\] from the equation

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.1O....... sqrt=1/2, 4sqrt=4/2=2. Got it?

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0cool, thank you :) much more sense

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.1My pleasure.... Don't forget to add a medal and be a fan. And definitely ask more questions. Will try my best to help!

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0which would leave me with \[4u^2+2(2u+1)+4u\] as the derivative?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1\[\sqrt[4]{u^2+u} \implies (u^2+u)^{1/4}\] \[\frac{ d }{ du } (u^2+u)^{1/4} = \frac{ 1 }{ 4 }(u^2+u)^{3/4} \times \frac{ d }{ du }(u^2+u)\] notice we apply the chain rule.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1So your derivative should be \[\frac{ 1 }{ 4 }(u^2+u)^{3/4} (2u+1)\]

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0dw:1438235581850:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Or you can put it as \[\frac{ (2u+1) }{ 4(u^2+u)^{3/4} }\]

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0please clarify the original notice @Astrophysics and I did 2 different problems

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1\(\color{blue}{\text{Originally Posted by}}\) @Carissa15 \[\sqrt[4]{u^2+u}\] \(\color{blue}{\text{End of Quote}}\) @triciaal

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0yes, that was the original that I needed to find the derivative of, however I struggle with taking the root when using derivatives.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Just remember your exponent rules if you have trouble with square roots \[\huge \sqrt[n]{x} \implies x^{1/n}\] then you can easily use power rule.

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.1@Astrophysics is here... he is really good.... He will take care..... @Carissa15

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0I see that now in the body but not in the original

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0Ok, thanks. Easier to work with powers

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0@Astrophysics did you see the original?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Alright, so everything is clear now? Another thing, the way they show derivatives for square root for example \[\frac{ d }{ dx } \sqrt{x} = \frac{ 1 }{ 2\sqrt{x} }\] notice they show it as such right? This is the same thing as \[\frac{ d }{ dx } x^{1/2} = \frac{ 1 }{ 2 }x^{1/2} = \frac{ 1 }{ 2x^{1/2} } = \frac{ 1 }{ 2\sqrt{x} }\] And yes I did @triciaal I could see your confusion, don't worry about it! :)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1So just mess around, and see what you get, that's the best way to learn!

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0I am not confused I read what was posted just as it was posted.

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0when using the power rule, you subtract by 1 (to the power) while multiple the rest of the relevant equation. But when you have fraction powers such as this?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Yes, it's the same, the power rule is as follow \[\frac{ d }{ dx } x^{n} = nx^{n1}\]

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0got it. Thank you everyone :)

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0got it. Thank you everyone :)
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