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))#