How do you find the derivative #y=(-3-2x+2x^3)/(-2-x^2)# using the quotient rule?

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Answer 1 · 2018-07-10T05:55:19

The answer is #=-(2(x^4+7x^2+3x-2))/(2+x^2)^2#

The derivative of a quotient is

#(u/v)'=(u'v-uv')/(v^2)#

Here, we have

#y=(-3-2x+2x^3)/(-2-x^2)#

#u=-3-2x+2x^3#, #=>#, #u'=-2+6x^2#

#v=-2-x^2#, #=>#, #v'=-2x#

Therefore,

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

#=(4+2x^2-12x^2-6x^4-6x-4x^2+4x^4)/(2+x^2)^2#

#=(4-6x-14x^2-2x^4)/(2+x^2)^2#

#=-(2(x^4+7x^2+3x-2))/(2+x^2)^2#