What is the derivative of #sqrt(x - 1)/sqrtx#?

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Answer 1 · 2016-01-27T05:05:03

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

To find the derivative of #sqrt(x-1)/sqrt(x)# we can try the following
#sqrt(x-1)/sqrt(x) = sqrt((x-1)/x)#
on simplifying further

#=sqrt(1-1/x)#

Let us use chain rule.

Let #y=sqrt(u)# and #u=1-1/x#

By chain rule

#dy/dx = dy/(du) xx (du)/dx#

#y=sqrt(u)#
#dy/(du) = 1/(2sqrt(u))#
Substituting back for #u# we get
#dy/(du) = 1/(2sqrt(1-1/x))#

Now we shall find our #(du)/dx#

#u=1-1/x#
#(du)/dx = 0- (-1/x^2)#
#(du)/dx = 1/x^2#

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

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