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just wondering how to proceed with this problem.

Calculus1
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Use the appropriate rule or combination of rules to find the derivative of the function defined below. y=\[(16x-4)\sqrt{8x-2}\]
i know i have to use my product rule, when i do that i get: (16x-4)(1/2)(8x-2)^-1/2 + (8x-2)^1/2 *16
u forgot chain rule for (8x-2)

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Other answers:

oh times 8 in there lol. i had that wrote down, lol just forgot to type it.
then u are correct
okay so that i got: \[\frac{ 4(16x-4) }{ \sqrt{8x-2} }+16\sqrt{8x-2}\]
absolutely correct!
would i take my 4 out of the numerator so i got 16(4x-1) ?
thats not much of a simplification, but u can !
is tehre a way to further simplify this> my answer isn't matching any of the answers on my assingment (this one is a multiple choice)
what are the options ? u can factor out 16 from both the terms..
a) 16- 4/squrt8x-2 B)16(squrt8x-2)+32 times 16x-4/squrt8x-2 c)16-16/squrt8x-2 d)64/squrt8x-2 e)32*2^1/2*x^1/2 -16*2^1/2+(16-4x)/squrt8x-2 f)8*csgn(4x-1)*2^1/2 g)16*squrt8x-2+4(16x-4/squrt8x-2) h)8(16x-4)(squrt8x-2)
ohh, u can cross-multiply to get the common denominator as \(\sqrt{}8x-2\)
so then i have\[4(16x-4)+16\sqrt{8x-2}(\sqrt{8x-2})\]
??
which is? \(\sqrt{8x-2}\sqrt{8x-2}=8x-2\)
i had 320x-16 as my numerator but that's not there as an answer either.
is the answer c?
checking, wait..your c option is weird...16-16=0
\[16-\frac{ 16 }{ \sqrt{8x-2} }\]
that's what it actually is, it's a little weird when i wrote it the other way lol..
so is that right tho?
My answer's the denom. cancel out! However the way your post the option too complicated to read !!!
how did your answer look? i can re-write them out.
\[a) 16-\frac{ 4 }{ \sqrt{8x-2} }\]
= 24 āˆš ( 8x -1)
\[b) 16\sqrt{8x-2}+32*\frac{ 16x-4 }{ \sqrt{8x-2} }\]
\[c)16-\frac{ 16 }{ 8x-2 }\]
\[d)\frac{ 64 }{ \sqrt{8x-2} }\]
\[e) 32*2^{\frac{ 1 }{ 2 }}*x^\frac{ 1 }{ 2 }-16*2^\frac{ 1 }{ 2 }+(16x-4)*\frac{ 2^\frac{ 1 }{ 2 } }{ x \frac{ 1 }{ 2 } }\]
\[f)8*csgn(4x-1)*2^\frac{ 1 }{ 2 }\]
are you sure?
\[g)16\sqrt{8x-2}+4\frac{ 16-4x }{ \sqrt{8x-2}}\]
no idea hah.
\[h)8(16-4x)(\sqrt{8x-2}\]
isn't g) option the same thing what we got ?
omg.. look at that.
hhaha so it is!!!
if you wanna check out my lastest/last question that would be great!
So it turns out that there's no need to simplified =)
i had the right answer all along haha.

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