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Mathematics 16 Online
OpenStudy (anonymous):

derivative of arcsin(cos(2x)) is this chain rule?

OpenStudy (anonymous):

Yup. Let cos(2x) = u, then d(arcsin(u))/du = ((1-x^2)^(-1)) * du' Then let v = 2x, then d(cos(v)) = -sin(v) v' It's just a double chain rule. Try it out.

OpenStudy (anonymous):

Minor correction, I meant ^(-1/2)

OpenStudy (anonymous):

i got -2sin(2x) * 1/(sqrt(1-(cos2x)^2)

OpenStudy (anonymous):

Let me work this through: \[ -2\frac{1}{\sqrt{1-cos(2x)^2}} \sin(2x) \]So you are correct :-) Congrats.

OpenStudy (anonymous):

how can I simplify the denominator?

OpenStudy (anonymous):

double angle?

OpenStudy (anonymous):

Hmm, I think you can do some algebra here (without double angles), but I just can't see it. Sorry, mate. :-)

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