What is the derivative of $y = arccsc (x/2)$ ?

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Answer 1 · 2016-06-30T22:51:17

$- 2/(x sqrt(x^2 - 4))$

if $y = csc^{-1} (x/2)$

then

$csc y = x/2$ [..... Which means that $color{red}{sin y = 2/x}$ ]

so

$D_x(csc y = x/2)$

$\implies - csc y \ cot y \ y ' = 1/2$

[ $D_z (csc z) = - csc z cot z$ is a well known derivative]

So we have
$y ' = 1/2 1/(- csc y \ cot y)$

$= - 1/2 sin y \ tan y$

the significance of the text in red is this:

enter image source here

because it should be clear that $tan y = 2/sqrt(x^2 - 4)$

so

$y ' = - 1/2 * 2/x * 2/sqrt(x^2 - 4)$

$= - 2/(x sqrt(x^2 - 4))$

What is the derivative of y = arccsc (x/2)y = arccsc (x/2)?