Asked by Jim
this is a question I don't understand:
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
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
Answers
Answered by
Steve
That'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
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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