Question #86235

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Answer 1 · 2017-11-08T16:52:16

Look below

Rewrite it as #arccos(1/x)#

now use the chain rule

#= -\frac{1}{sqrt(1-(1/x)^2}) times \frac{d}{dx}[1/x]#

#=d/dx[1/x]=\frac{d/dx[x]}{x^2}#

now subsitute the 1 with the function

#= -\frac{\frac{d/dx[x]}{x^2}}{sqrt(1-(1/x)^2})#

move the #x^2# to the denominator

#= \frac{1}{sqrt(1-(1/x)^2)x^2#

Answer 2 · 2017-11-09T15:23:06

#1/(|x|sqrt(x^2-1))#

let

#y=cos^(-1)(1/x)#

#=>1/x=cosy#

differentiate #wrt" "x#

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

#(dy)/(dx)=1/(x^2siny)#

#=1/(x^2(1-cos^2y)#

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

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

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

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