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Calculus1 14 Online
OpenStudy (anonymous):

just wondering how to proceed with this problem.

OpenStudy (anonymous):

Use the appropriate rule or combination of rules to find the derivative of the function defined below. y=\[(16x-4)\sqrt{8x-2}\]

OpenStudy (anonymous):

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

hartnn (hartnn):

u forgot chain rule for (8x-2)

OpenStudy (anonymous):

oh times 8 in there lol. i had that wrote down, lol just forgot to type it.

hartnn (hartnn):

then u are correct

OpenStudy (anonymous):

okay so that i got: \[\frac{ 4(16x-4) }{ \sqrt{8x-2} }+16\sqrt{8x-2}\]

hartnn (hartnn):

absolutely correct!

OpenStudy (anonymous):

would i take my 4 out of the numerator so i got 16(4x-1) ?

hartnn (hartnn):

thats not much of a simplification, but u can !

OpenStudy (anonymous):

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)

hartnn (hartnn):

what are the options ? u can factor out 16 from both the terms..

OpenStudy (anonymous):

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)

hartnn (hartnn):

ohh, u can cross-multiply to get the common denominator as \(\sqrt{}8x-2\)

OpenStudy (anonymous):

so then i have\[4(16x-4)+16\sqrt{8x-2}(\sqrt{8x-2})\]

OpenStudy (anonymous):

??

hartnn (hartnn):

which is? \(\sqrt{8x-2}\sqrt{8x-2}=8x-2\)

OpenStudy (anonymous):

i had 320x-16 as my numerator but that's not there as an answer either.

OpenStudy (anonymous):

is the answer c?

hartnn (hartnn):

checking, wait..your c option is weird...16-16=0

OpenStudy (anonymous):

\[16-\frac{ 16 }{ \sqrt{8x-2} }\]

OpenStudy (anonymous):

that's what it actually is, it's a little weird when i wrote it the other way lol..

OpenStudy (anonymous):

so is that right tho?

OpenStudy (anonymous):

My answer's the denom. cancel out! However the way your post the option too complicated to read !!!

OpenStudy (anonymous):

how did your answer look? i can re-write them out.

OpenStudy (anonymous):

\[a) 16-\frac{ 4 }{ \sqrt{8x-2} }\]

OpenStudy (anonymous):

= 24 √ ( 8x -1)

OpenStudy (anonymous):

\[b) 16\sqrt{8x-2}+32*\frac{ 16x-4 }{ \sqrt{8x-2} }\]

OpenStudy (anonymous):

\[c)16-\frac{ 16 }{ 8x-2 }\]

OpenStudy (anonymous):

\[d)\frac{ 64 }{ \sqrt{8x-2} }\]

OpenStudy (anonymous):

\[e) 32*2^{\frac{ 1 }{ 2 }}*x^\frac{ 1 }{ 2 }-16*2^\frac{ 1 }{ 2 }+(16x-4)*\frac{ 2^\frac{ 1 }{ 2 } }{ x \frac{ 1 }{ 2 } }\]

OpenStudy (anonymous):

\[f)8*csgn(4x-1)*2^\frac{ 1 }{ 2 }\]

OpenStudy (anonymous):

are you sure?

OpenStudy (anonymous):

\[g)16\sqrt{8x-2}+4\frac{ 16-4x }{ \sqrt{8x-2}}\]

OpenStudy (anonymous):

no idea hah.

OpenStudy (anonymous):

\[h)8(16-4x)(\sqrt{8x-2}\]

hartnn (hartnn):

isn't g) option the same thing what we got ?

OpenStudy (anonymous):

omg.. look at that.

OpenStudy (anonymous):

hhaha so it is!!!

OpenStudy (anonymous):

if you wanna check out my lastest/last question that would be great!

OpenStudy (anonymous):

So it turns out that there's no need to simplified =)

OpenStudy (anonymous):

i had the right answer all along haha.

OpenStudy (anonymous):

http://openstudy.com/study#/updates/50902c28e4b043900ad9134d please : )

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