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How do you find the first and second derivatives of #y= (6x^2+4)/ (2x+2)# using the quotient rule?

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Jun 17, 2018

Answer:

#y'=(2x^2+6x-2)/(x+1)^2#
#y''=10/(x+1)^3#

Explanation:

Using the Quotient rule we get

#y'=(6x(x+1)-(3x^2+2))/(x+1)^2#
which simplifyes to

#y'=(3x^2+6x-2)/(x+1)^2#

Again by the Quotient rule we get

#y''=((6x+6)(x+1)^2-(3x^2+6x-2)(2(x+1)))/(x+1)^3#
Simplifying we obtain
#y''=10/(x+1)^3#

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