What is the derivative of this function #y=sin^-1(1/x)#?
1 Answer
Jul 13, 2016
Explanation:
#color(orange)"Reminder" d/dx(sin^-1x)=1/(sqrt(1-x^2)# here, however, x =
#1/x# Differentiate using the
#color(blue)" chain rule combined with power rule"#
#color(orange)" Chain rule"#
#color(red)(|bar(ul(color(white)(a/a)color(black)(d/dx(f(g(x)))=f'(g(x))g'(x))color(white)(a/a)|)))........ (A)#
#"---------------------------------------------------------------"#
#f(g(x))=sin^-1(1/x)rArrf'(g(x))=1/(sqrt(1-(1/x)^2)# and
#g(x)=1/x=x^-1rArrg'(x)=-x^-2=-1/x^2#
#"-----------------------------------------------------------------"#
Substitute these values into (A)
#=1/(sqrt(1-1/x^2))xx-1/x^2#
#=(-1)/(x^2sqrt(1/x^2(x^2-1))#
#=(-1)/(x^2xx1/xsqrt(x^2-1)#
#rArrd/dx(sin^-1(1/x))=(-1)/(xsqrt(x^2-1)#