How do you find the derivative of #y= (2x)/ sqrt(x - 1)#?

1 Answer
Ratnaker Mehta
Jun 29, 2016

#dy/dx=(x-2)/{(x-1)sqrt(x-1)}.#

Explanation:

Method I uses the Quotient Rule for Diffn. : #(u/v)'=(v*u'-u*v')/v^2#.

Let #y=u/v=(2x)/sqrt(x-1).#

#u=2x rArr u'=2.#
#v=sqrt(x-1)=(x-1)^(1/2) rArr v'=1/2*(x-1)^(1/2-1)*d/dx(x-1)=1/2(x-1)^(-1/2)*1=1/(2sqrt(x-1)#

Sub.ing, these values in the Quotient Rule, we have,

#dy/dx={sqrt(x-1)*2-(2x)/(2sqrt(x-1)}}/(x-1)={2sqrt(x-1)-x/sqrt(x-1)}/(x-1)=(2x-2-x)/{(x-1)*sqrt(x-1)}=(x-2)/{(x-1)sqrt(x-1)}.#

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