Math (Calculus)
posted by Jim .
without any simplification, find the derivative of
h(x) Arcsin^4 (3x1) / (3)^1/2 (square root)
I end up with h'(x) =
12Arcsin^3 (3x1)sqr(3) / 3 (squr(1(3x)^2))

I'm just not sure if I have to do the product of both Arcsin^4 (3x1) and then do the quotient rule with sqr(3), or do it the way I just did, considering Arcsin^4 (3x1) as one and do the quotient rule right away
Thank you 
You are correct, except that in the denominator, that should be (3x1)^2, not (3x)^2
If you simplify the radicand by expanding and folding in the sqrt(3) from the numerator, you end up with
4Arcsin^3 (3x1) / sqrt(2x3x^2)
There is no quotient here. sqrt(3) is just a constant, and you just have u^4 where u = arcsin(v) and v = 3x1; use the regular power rule, and the chain rule twice.