Find the differential of y in the function: y=(x^2-4)/(x-3) ?

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Answer 1 · 2018-03-04T08:40:38

$dy/dx=(x^2-6x+4)/(x-3)^2$

We want to find the derivative of

$y=(x^2-4)/(x-3)$

Use the quotient rule, if $y=f/g$

then $dy/dx=(f'*g-f*g')/g^2$ , thus

By the quotient rule

$dy/dx=(2x(x-3)-(x^2-4))/(x-3)^2$

$=(2x^2-6x-x^2+4)/(x-3)^2$

$=(x^2-6x+4)/(x-3)^2$