how do I differentiate 4sqrt(u^2+u) ?

- Carissa15

how do I differentiate 4sqrt(u^2+u) ?

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- arindameducationusc

with respect to u?

- anonymous

WELCOME TO OPENSTUDY!

- anonymous

i will help ya :)

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## More answers

- Carissa15

Oh yes, sorry. It says Differentiate f(u) I have not yet differentiated using roots yet

- anonymous

lol fu

- anonymous

ok so first:

- Carissa15

\[\sqrt[4]{u^2+u}\]

- anonymous

so first of all do you know how to simplify the equation?

- arindameducationusc

See this......

##### 1 Attachment

- arindameducationusc

@Carissa15 you there?

- Carissa15

yes, that makes sense so far

- Carissa15

I thought that you add 1/2 to "remove" the root from the equation but then I get lost..

- arindameducationusc

??

- 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

O....... sqrt=1/2, 4sqrt=4/2=2. Got it?

- Carissa15

cool, thank you :-) much more sense

- 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

which would leave me with \[4u^2+2(2u+1)+4u\] as the derivative?

- 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

So your derivative should be \[\frac{ 1 }{ 4 }(u^2+u)^{-3/4} (2u+1)\]

- triciaal

|dw:1438235581850:dw|

- Astrophysics

Or you can put it as \[\frac{ (2u+1) }{ 4(u^2+u)^{3/4} }\]

- triciaal

please clarify the original notice @Astrophysics and I did 2 different problems

- Astrophysics

\(\color{blue}{\text{Originally Posted by}}\) @Carissa15
\[\sqrt[4]{u^2+u}\]
\(\color{blue}{\text{End of Quote}}\)
@triciaal

- 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

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

@Astrophysics is here... he is really good.... He will take care..... @Carissa15

- triciaal

I see that now in the body but not in the original

- Carissa15

Ok, thanks. Easier to work with powers

- triciaal

@Astrophysics did you see the original?

- 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

So just mess around, and see what you get, that's the best way to learn!

- triciaal

I am not confused I read what was posted just as it was posted.

- 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

Yes, it's the same, the power rule is as follow \[\frac{ d }{ dx } x^{n} = nx^{n-1}\]

- Carissa15

got it. Thank you everyone :-)

- Astrophysics

Np

- Carissa15

got it. Thank you everyone :-)

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