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Find the derivative of the function y = sin^[-1](2x+1)

Mathematics
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i got 2/sqrt(1-(2x+1)^2) but apparently that ain't right
did you try solving out the (2x+1)^2?
|dw:1333934423544:dw|

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Other answers:

expand the radicand and factor out the 4 to get the 2's to cancel... the you'll get y'=1/sqrt(-x^2-x)
uh.. no...? (2x+1)^2 is 4x^2 + 4x + 1. right. so how does that work.
crap
what about -2sin^-2(2x+1)*cos(2x+1)
my trig is no good man just show me the way any way just show me so i get it
didnt use trig. just differentiation. treat it as (sin(2x+1))^-1 . bring down the -1, raise whole thing to -2. thats one part of the answer. Then multiply that by the derivative of the inside. The inside (sin(2x+1)) has a derivative of cos(2x+1). Thats the second part. The last part is the derivative of the inside of that. derivative of 2x+1 is 2. Theres the three parts. Multiply them all together. better?
well ok so it's -2 * 1/sin(2x+1)^2 * cos(2x+1)... so it's -2cos(2x+1)/sin(2x+1)^2..... soooooo.........
do i have that right so far? and then what?
thats it i believe. is it not right?
no dpalnc had it it's y'=1/sqrt(-x^2-x) in the textbook
so.... yeah oy
okay, http://www.wolframalpha.com/input/?i=derivative+of++y+%3D+sin%5E-1%282x%2B1%29 i just looked it up. They use trig substitution and turn it into variables. look at the steps.
what. it just gives me the same answer i got. where do i go from there.
it doesn't bring you to the fluttering end solution that it shows it just brings me to 2/sqrt(1-(2x+1)^2 which is what i got anyway
nvm
arccos x just means the inverse cosine function right like cos^-1 (x) ??
|dw:1333946189893:dw|
ok thanks but that last q was unrelated
take out the 4... so it will be 2 outside.... the radical...

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