Question

cot(arcsin 3x)

Answer 1

What you need to do for this is first know that cot is equal to cos/sin, so
\[\frac{ \cos(\arcsin3x) }{\sin(\arcsin3x) }\]

Answer 2

Next step, you know that sin is the inverse of arcsin, so they cancel eachother out.
This gives us
\[\frac{ \cos(arcsom3x) }{ 3x }\]

Answer 3

Lastly, cos(arcsin x) = sin(arccos x) = √(1 − x²)

Multiplied together is effectively the same as squaring... the radical disappears except the range on 1-x² is limited to the positive values. You don't need to know this, simply memorize the identities.

So we have
\[\frac{ \sqrt{1-9x^2} }{ 9x }\]

Answer 4

Peace

Answer 5

sqrt(9 - 9x^2)/9x