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

1. arindameducationusc

with respect to u?

2. anonymous

WELCOME TO OPENSTUDY!

3. anonymous

i will help ya :)

4. Carissa15

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

5. anonymous

lol fu

6. anonymous

ok so first:

7. Carissa15

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

8. anonymous

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

9. arindameducationusc

See this......

10. arindameducationusc

@Carissa15 you there?

11. Carissa15

yes, that makes sense so far

12. Carissa15

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

13. arindameducationusc

??

14. 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

15. arindameducationusc

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

16. Carissa15

cool, thank you :-) much more sense

17. 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!

18. Carissa15

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

19. 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.

20. Astrophysics

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

21. triciaal

|dw:1438235581850:dw|

22. Astrophysics

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

23. triciaal

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

24. Astrophysics

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

25. Carissa15

yes, that was the original that I needed to find the derivative of, however I struggle with taking the root when using derivatives.

26. 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.

27. arindameducationusc

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

28. triciaal

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

29. Carissa15

Ok, thanks. Easier to work with powers

30. triciaal

@Astrophysics did you see the original?

31. 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! :)

32. Astrophysics

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

33. triciaal

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

34. 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?

35. Astrophysics

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

36. Carissa15

got it. Thank you everyone :-)

37. Astrophysics

Np

38. Carissa15

got it. Thank you everyone :-)

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