Question
Find $\frac{dy}{dx}$, $y=\frac{2x+3}{3x-2}$.
✓
Answer
Using Quotient Rule : $\frac{v(du/dx) - u(dv/dx)}{v^2}$
$\frac{(3x-2)(2) - (2x+3)(3)}{(3x-2)^2}$
$= \frac{6x - 4 - 6x - 9}{(3x-2)^2}$
$ = \frac{-13}{(3x-2)^2}$.
$\frac{(3x-2)(2) - (2x+3)(3)}{(3x-2)^2}$
$= \frac{6x - 4 - 6x - 9}{(3x-2)^2}$
$ = \frac{-13}{(3x-2)^2}$.
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