What is the derivative of this function $y=csc^-1(x/2)$ ?

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Answer 1 · 2016-12-23T05:45:59

The answer is $=-2/(x^2sqrt(1-4/x^2))$

We use

$cos^2theta+sin^2theta=1$

and $(x^n)'=nx^(n-1)$

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

$cscy=x/2$

$1/siny=x/2$

$siny=2/x$

$(siny)'=(2/x)'$

$cosydy/dx=-2/x^2$

$dy/dx=-2/(x^2cosy)$

$cosy=sqrt(1-sin^2y)=sqrt(1-4/x^2)$

Therefore,

$dy/dx=-2/(x^2sqrt(1-4/x^2))$

What is the derivative of this function y=csc^-1(x/2)y=csc^-1(x/2)?