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February 28, 2015

February 28, 2015

Posted by **Jim** on Tuesday, November 15, 2011 at 1:33am.

Demonstrate the following derivative rule:

(Arccsc(u))' = (-1/u(u^2-1)^1/2) * u'

Where u = g(x)

Do they want me to start with (Arccsc(u)) and get to (-1/u(u^2-1)^1/2) * u' ?

Thank you

- Math (Calculus) -
**Steve**, Tuesday, November 15, 2011 at 5:02pmThat's the idea, but you have to go to it backwards, in a way

if y = arccsc(x) then x = csc(y)

dx/dy = -cscy ctny

but ctny = sqrt(csc^2 y - 1)

dx/dy = -x sqrt(x^2 - 1)

dy/dx = -1/[x * sqrt(x^2-1)]

If we have y = arccsc(u) then the chain rule says we have

dy/dx = dy/du * du/dx = -1/[u * sqrt(u^2-1)]

du/dx

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