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Mathematics 16 Online
OpenStudy (carissa15):

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

OpenStudy (arindameducationusc):

with respect to u?

OpenStudy (anonymous):

WELCOME TO OPENSTUDY!

OpenStudy (anonymous):

i will help ya :)

OpenStudy (carissa15):

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

OpenStudy (anonymous):

lol fu

OpenStudy (anonymous):

ok so first:

OpenStudy (carissa15):

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

OpenStudy (anonymous):

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

OpenStudy (arindameducationusc):

See this......

OpenStudy (arindameducationusc):

@Carissa15 you there?

OpenStudy (carissa15):

yes, that makes sense so far

OpenStudy (carissa15):

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

OpenStudy (arindameducationusc):

??

OpenStudy (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

OpenStudy (arindameducationusc):

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

OpenStudy (carissa15):

cool, thank you :-) much more sense

OpenStudy (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!

OpenStudy (carissa15):

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

OpenStudy (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.

OpenStudy (astrophysics):

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

OpenStudy (triciaal):

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