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#