What is the derivative of this function #y=xcsc^-1x#?

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Answer 1 · 2016-10-23T19:30:17

Use the product rule and the derivative of #csc^-1x#

The derivative of #csc^-1x# depends on how you've defined it.

I use #y = csc^-1x# if and only if #cscy=x# and #- pi/2 < y < pi/2# with #y != 0#.

With this definition #d/dx(csc^-1 x ) = -1/(absx sqrt(x^2-1))#

For #f(x) = x csc^-1x#, we get

#f'(x) = csc^-1x - x/(absx sqrt(x^2-1))#

Alternatively

If you allow #0 < y < pi/2# or #pi < y < (3pi)/2#, then the absolute value is lost and we have

#d/dx(csc^-1 x ) = -1/(x sqrt(x^2-1))#

In this case

#f'(x) = csc^-1x - 1/sqrt(x^2-1)#