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Math (Calculus)

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without any simplification, find the derivative of

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) -

    I'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) -

    You 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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