How do you differentiate $y=(3x)/(x^2-3)$ ?

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Answer 1 · 2016-12-04T15:53:40

The derivative is $dy/dx=(-3(3+x^2))/(x^2-3)^2$

This thequotient of two functions

The derivative is $(u/v)'=(u'v-uv')/v^2$

Here. $u=3x$ , $u'=3$

and $v=x^2-3$ , $v'=2x$

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

$=(3x^2-9-6x^2)/(x^2-3)^2$

$=(-9-3x^2)/(x^2-3)^2$

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

How do you differentiate y=(3x)/(x^2-3)y=(3x)/(x^2-3)?