Carissa15
  • Carissa15
how do I differentiate 4sqrt(u^2+u) ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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arindameducationusc
  • arindameducationusc
with respect to u?
anonymous
  • anonymous
WELCOME TO OPENSTUDY!
anonymous
  • anonymous
i will help ya :)

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Carissa15
  • Carissa15
Oh yes, sorry. It says Differentiate f(u) I have not yet differentiated using roots yet
anonymous
  • anonymous
lol fu
anonymous
  • anonymous
ok so first:
Carissa15
  • Carissa15
\[\sqrt[4]{u^2+u}\]
anonymous
  • anonymous
so first of all do you know how to simplify the equation?
arindameducationusc
  • arindameducationusc
See this......
1 Attachment
arindameducationusc
  • arindameducationusc
@Carissa15 you there?
Carissa15
  • Carissa15
yes, that makes sense so far
Carissa15
  • Carissa15
I thought that you add 1/2 to "remove" the root from the equation but then I get lost..
arindameducationusc
  • arindameducationusc
??
Carissa15
  • Carissa15
So 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
  • arindameducationusc
O....... sqrt=1/2, 4sqrt=4/2=2. Got it?
Carissa15
  • Carissa15
cool, thank you :-) much more sense
arindameducationusc
  • arindameducationusc
My pleasure.... Don't forget to add a medal and be a fan. And definitely ask more questions. Will try my best to help!
Carissa15
  • Carissa15
which would leave me with \[4u^2+2(2u+1)+4u\] as the derivative?
Astrophysics
  • Astrophysics
\[\sqrt[4]{u^2+u} \implies (u^2+u)^{1/4}\] \[\frac{ d }{ du } (u^2+u)^{1/4} = \frac{ 1 }{ 4 }(u^2+u)^{-3/4} \times \frac{ d }{ du }(u^2+u)\] notice we apply the chain rule.
Astrophysics
  • Astrophysics
So your derivative should be \[\frac{ 1 }{ 4 }(u^2+u)^{-3/4} (2u+1)\]
triciaal
  • triciaal
|dw:1438235581850:dw|
Astrophysics
  • Astrophysics
Or you can put it as \[\frac{ (2u+1) }{ 4(u^2+u)^{3/4} }\]
triciaal
  • triciaal
please clarify the original notice @Astrophysics and I did 2 different problems
Astrophysics
  • Astrophysics
\(\color{blue}{\text{Originally Posted by}}\) @Carissa15 \[\sqrt[4]{u^2+u}\] \(\color{blue}{\text{End of Quote}}\) @triciaal
Carissa15
  • Carissa15
yes, that was the original that I needed to find the derivative of, however I struggle with taking the root when using derivatives.
Astrophysics
  • Astrophysics
Just 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
  • arindameducationusc
@Astrophysics is here... he is really good.... He will take care..... @Carissa15
triciaal
  • triciaal
I see that now in the body but not in the original
Carissa15
  • Carissa15
Ok, thanks. Easier to work with powers
triciaal
  • triciaal
@Astrophysics did you see the original?
Astrophysics
  • Astrophysics
Alright, 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
  • Astrophysics
So just mess around, and see what you get, that's the best way to learn!
triciaal
  • triciaal
I am not confused I read what was posted just as it was posted.
Carissa15
  • Carissa15
when 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
  • Astrophysics
Yes, it's the same, the power rule is as follow \[\frac{ d }{ dx } x^{n} = nx^{n-1}\]
Carissa15
  • Carissa15
got it. Thank you everyone :-)
Astrophysics
  • Astrophysics
Np
Carissa15
  • Carissa15
got it. Thank you everyone :-)

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