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Mathematics 23 Online
OpenStudy (anonymous):

can someone explain this simple derivative to me please: sort(6x)

OpenStudy (anonymous):

isn't it 1/2(6x)^(-1/2)

OpenStudy (anonymous):

how does it go from there to the answer: sqrt(6)/2 sqrt(x)

OpenStudy (psymon):

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.

OpenStudy (psymon):

\[\sqrt{6x} = \frac{ 1 }{ 2 }(6x)^{\frac{ -1 }{ 2 }}(6)\]

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

im no sure about the chain rule its confusing

OpenStudy (anonymous):

i know its: [f(g(x))]'= f'(g(x))*g'(x)

OpenStudy (psymon):

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?

OpenStudy (anonymous):

no

OpenStudy (anonymous):

oh wait, i think so a while ago.

OpenStudy (anonymous):

let me open

OpenStudy (psymon):

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.

OpenStudy (anonymous):

does it always work?

OpenStudy (psymon):

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.

OpenStudy (anonymous):

thanks im going to read it right now

OpenStudy (psymon):

Yeah, just let me know.

OpenStudy (anonymous):

ugh i gave it a shot. wrong. :(

OpenStudy (psymon):

How'd it become 1/4?

OpenStudy (anonymous):

i did 1/2 *-1/2

OpenStudy (psymon):

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.

OpenStudy (anonymous):

huh?

OpenStudy (psymon):

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.

OpenStudy (anonymous):

down in the denominator?

OpenStudy (anonymous):

can you show me the steps to the answer? just want to see how you did it so i can ask

OpenStudy (psymon):

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 \[\frac{ d }{ dx }k ^{n} = nk ^{n-1}\], 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: \[\frac{ 3 }{ \sqrt{6x} }\]

OpenStudy (anonymous):

oh the answer is sqrt (6)/2 sqrt(x)

OpenStudy (psymon):

It's the same thing actually xD

OpenStudy (psymon):

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: \[\frac{ \sqrt{6} }{ 2\sqrt{x} }\]

OpenStudy (anonymous):

alrigght thanks ! :)))

OpenStudy (psymon):

Mhm, np ^_^ Hope that made sense xD

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