Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
mathcalculus
Group Title
can someone explain this simple derivative to me please: sort(6x)
 11 months ago
 11 months ago
mathcalculus Group Title
can someone explain this simple derivative to me please: sort(6x)
 11 months ago
 11 months ago

This Question is Closed

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
isn't it 1/2(6x)^(1/2)
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
how does it go from there to the answer: sqrt(6)/2 sqrt(x)
 11 months ago

Psymon Group TitleBest ResponseYou've already chosen the best response.1
It's chain rule so you also need to multiply by the derivative of the inside. So it would be what you put, but then multiply by 6.
 11 months ago

Psymon Group TitleBest ResponseYou've already chosen the best response.1
\[\sqrt{6x} = \frac{ 1 }{ 2 }(6x)^{\frac{ 1 }{ 2 }}(6)\]
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
im no sure about the chain rule its confusing
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i know its: [f(g(x))]'= f'(g(x))*g'(x)
 11 months ago

Psymon Group TitleBest ResponseYou've already chosen the best response.1
Well, I have a unique way of showing it, so maybe it will help maybe not. Were you one of the ones I sent the derivative files to?
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
oh wait, i think so a while ago.
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
let me open
 11 months ago

Psymon Group TitleBest ResponseYou've already chosen the best response.1
Ah xD Yeah, one of them had the chain rule in it. I teach it in a unique way, so maybe it helps maybe it doesn't.
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
does it always work?
 11 months ago

Psymon Group TitleBest ResponseYou've already chosen the best response.1
Yeah. As long as you recognize what layers you have then yeah. Chain rule is multiplying the derivatives of each layer you have, making sure to not disturb what was originally inside of each layer.
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
thanks im going to read it right now
 11 months ago

Psymon Group TitleBest ResponseYou've already chosen the best response.1
Yeah, just let me know.
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
ugh i gave it a shot. wrong. :(
 11 months ago

Psymon Group TitleBest ResponseYou've already chosen the best response.1
How'd it become 1/4?
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i did 1/2 *1/2
 11 months ago

Psymon Group TitleBest ResponseYou've already chosen the best response.1
Well, you bring the power down and then you made the power become 1/2. You're done with that layer, now you just go to the inner layer. \[\frac{ 1 }{ 2 }()^{\frac{ 1 }{ 2 }}\] Thats it, no more to that part.
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
huh?
 11 months ago

Psymon Group TitleBest ResponseYou've already chosen the best response.1
Well, when you do the outer layer. All you do is bring the power down as a multiplication then lower the power by 1. After that you are done with that layer. There's no other 1/2 to multiply by.
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
down in the denominator?
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
can you show me the steps to the answer? just want to see how you did it so i can ask
 11 months ago

Psymon Group TitleBest ResponseYou've already chosen the best response.1
Alright, so we have two layers. The first layer is simply ( )^1/2 and the inner layer 6x. So the chain rule says that we take the derivative of each layer and multiply the results. So following the normal derivative rule of \[\frac{ d }{ dx }k ^{n} = nk ^{n1}\], I'll do that with the first layer. \[\frac{ 1 }{ 2 }()^{\frac{ 1 }{ 2 }}\]I just left the inner part blank for now, but that is the derivative of the first layer. Now I do the second layer, which is just 6x. So the derivativeof 6x is simply 6. So now that I have the derivative of both layers, I now multiply both of these derivatives \[\frac{ 1 }{ 2 }()^{\frac{ 1 }{ 2 }}*(6)\] This of course becomes: \[\frac{ 3 }{ ()^{\frac{ 1 }{ 2 }}}\] Now all that is left to do is plug back in what was originally inside of that layer, giving us: \[\frac{ 3 }{ \sqrt{6x} }\]
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
oh the answer is sqrt (6)/2 sqrt(x)
 11 months ago

Psymon Group TitleBest ResponseYou've already chosen the best response.1
It's the same thing actually xD
 11 months ago

Psymon Group TitleBest ResponseYou've already chosen the best response.1
I'll show ya why: \[\frac{ 3 }{ \sqrt{6x} }=\frac{ 3 }{ (\sqrt{6})(\sqrt{x)} }\]Now multiply top and bottom by sqrt(6) \[\frac{ 3(\sqrt{6)} }{ (\sqrt{6})(\sqrt{x})(\sqrt{6}) }\]This then becomes finally: \[\frac{ \sqrt{6} }{ 2\sqrt{x} }\]
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
alrigght thanks ! :)))
 11 months ago

Psymon Group TitleBest ResponseYou've already chosen the best response.1
Mhm, np ^_^ Hope that made sense xD
 11 months ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.