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Posted by **Jim** on Tuesday, November 15, 2011 at 11:45am.

h(x) Arcsin^4 (3x-1) / (3)^1/2 (square root)

I end up with h'(x) =

12Arcsin^3 (3x-1)sqr(3) / 3 (squr(1-(3-x)^2))

- Math (Calculus) -
**Jim**, Tuesday, November 15, 2011 at 11:46amI'm just not sure if I have to do the product of both Arcsin^4 (3x-1) and then do the quotient rule with sqr(3), or do it the way I just did, considering Arcsin^4 (3x-1) as one and do the quotient rule right away

Thank you

- Math (Calculus) -
**Steve**, Tuesday, November 15, 2011 at 5:20pmYou are correct, except that in the denominator, that should be (3x-1)^2, not (3-x)^2

If you simplify the radicand by expanding and folding in the sqrt(3) from the numerator, you end up with

4Arcsin^3 (3x-1) / sqrt(2x-3x^2)

There is no quotient here. sqrt(3) is just a constant, and you just have u^4 where u = arcsin(v) and v = 3x-1; use the regular power rule, and the chain rule twice.

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