Asked by June
Can someone help me with finding the derivative of this function?
y = -( csc x)2 (cos-1(1-x2))
y = -( csc x)2 (cos-1(1-x2))
Answers
Answered by
Steve
product rule and chain rule -- if
y = uv
y' = u'v + uv'
Here u=csc^2 x and v = arccos(1-x^2)
y' = -2 cscx cscx cotx arccos(1-x^2) + csc^2 x 1/√(1-(1-x^2)^2) * -2x
kind of messy, no? well, recall that if cos u = 1-x^2, sin u = x, so we really have
y = -csc^2(x) arcsin(x)
y' = csc^2(x) (2arcsin(x)cot(x) - 1/√(1-x^2))
y = uv
y' = u'v + uv'
Here u=csc^2 x and v = arccos(1-x^2)
y' = -2 cscx cscx cotx arccos(1-x^2) + csc^2 x 1/√(1-(1-x^2)^2) * -2x
kind of messy, no? well, recall that if cos u = 1-x^2, sin u = x, so we really have
y = -csc^2(x) arcsin(x)
y' = csc^2(x) (2arcsin(x)cot(x) - 1/√(1-x^2))
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