What is the derivative of #arcsin(1/x)#?

1 Answer
Shiva Prakash M V
Feb 16, 2018

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

Explanation:

To find #d/dx(arcsin(1/x))#
#arcsin(1/x)=sin^-1(1/x)#
Let #y=sin^=1(1/x)#
Let #u=1/x#
#(du)/dx=(-1)/x^2#
#y=sin^-1u#
#(dy)/dx=1/sqrt(1-u^2)(du)/dx#
Substituting the values of #u and (du)/dx#
we have

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

Simplifying

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

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

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