Question #53041

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Answer 1 · 2017-10-17T07:49:36

#dy/dx=(-1)/(x^2sqrt(1-1/x^2)# or #dy/dx=(-1)/(xsqrt(x^2-1)#

We know the derivative of #(sin^-1x)# is #(1/sqrt(1-x^2))#

Therefore,

#y=sin^1(1/x)#

We will differentiate both sides using the chain rule.

#d/dxy=d/dx(sin^-1(1/x))#

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

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

#dy/dx1/sqrt(1-1/x^2)((-1)/x^2)#

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