What is the derivative of #y=arccos(1/x)#?

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Answer 1 · 2015-04-20T18:22:03

First find the value of siny because siny is going to appear when you use implicit differentiation to find the derivative of the function.

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Now differentiate and get the value of #dy/dx#.

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Answer 2 · 2015-04-20T18:45:50

The answer is: #y'=1/(xsqrt(x^2-1))#.

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

#=1/(x^2sqrt((x^2-1)/x^2))=1/(x^2*1/x*sqrt(x^2-1))=#

#=1/(xsqrt(x^2-1))#.