How do you differentiate #y= sqrt(x/(x-9))#?

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Answer 1 · 2017-05-02T17:01:19

#(dy)/(dx)=-9/(2(x-9)sqrt(x(x-9))#

Quotient rule states if #y(x)=(g(x))/(h(x))#

then #(dy)/(dx)=((dg)/(dx)xxh(x)-(dh)/(dx)xxg(x))/(h(x))^2#

Hence as #y+sqrt(x/(x-9))=sqrtx/sqrt(x-9)#

So here #g(x)=sqrtx# and #(dg)/(dx)=1/(2sqrtx)#

and #h(x)=sqrt(x-9)# and #(dh)/(dx)=1/(2sqrt(x-9))#

#(dy)/(dx)=(1/(2sqrtx)xxsqrt(x-9)-1/(2sqrt(x-9))xxsqrtx)/(x-9)#

#=((x-9)/2-x/2)/((x-9)sqrt(x(x-9))#

#=-9/(2(x-9)sqrt(x(x-9))#