# For y=\frac{3x+5}{x-3}, find the instantaneous rate of change when x=2

For $y=\frac{3x+5}{x-3}$, find the instantaneous rate of change when x=2
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Step 1
$y=\frac{3x+5}{x-3}$
$\frac{dy}{dx}=\frac{\left(x-3\right)\frac{d}{dx}\left(3x+5\right)-\left(3x+5\right)\frac{d}{dx}\left(x-3\right)}{{\left(x-3\right)}^{2}}$
Step 2
$\frac{dy}{dx}=\frac{\left(x-3\right)3-\left(3x+5\right)}{{\left(x-3\right)}^{2}}$
$\frac{dy}{dx}=\frac{3x-9-3x-5}{{\left(x-3\right)}^{2}}$
$\frac{dy}{dx}=\frac{-14}{{\left(x-3\right)}^{2}}$
${\left(\frac{dy}{dx}\right)}_{x=2}=\frac{-14}{{\left(2-3\right)}^{2}}$
${\left(\frac{dy}{dx}\right)}_{x=2}=-14$
So instantaneous rate of change =-14