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 $(dh)/(dx)=1/(2sqrt(x-9))$

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

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

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

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