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differentiate the stuff inside the arcsin, then multiply that by 7/sqrt(1-(stuff inside arcsin)^2)

lol he didn't give you the answer, he was giving you a way to solve the problem -_-

To start, do you know this derivative?\[\large y=\arcsin x \qquad \qquad y'=?\]

i know it's something like
1/sqrt(1-u)^2

|dw:1368923334862:dw|

that's what I was saying

Yes that sounds right bee, \[\large y=\arcsin x \qquad \qquad y'=\frac{1}{\sqrt{1-x^2}}\]

@klllerbee look at my first post...............

I said sqrt.....meaning square root

sqrt(1-(stuff inside arcsin)^2)

\[7/\sqrt(1-(10x^2+15x-9))^2\]

ok, cool. Thanks guys

How do i give you metals for helping :D
i'm new here

no, that's not the final answer.......

still got to multiply it by the derivative

oh
so the final answer is 7x(20x+15)/sqrt(1-(10x^2+15x-9)^2

yeah.....ze medal?

where is the metal button?

:( i don't know what i'm doing with this software