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Y=(3x-2)/(sqrt(2x+1)) differentiate

I have the quotient rule set up but I'm having trouble simplifying

1 Answer

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y=(3x-2)/(√(2x+1))=(3x-2)/((2x+1)^(1/2))

The quotient rule gives


y'=((2x+1)^(1/2)(3x-2)'-(3x-2)\left((2x+1)^(1/2)\right)')/(\left((2x+1)^(1/2)\right)^2)=(3(2x+1)^(1/2)-(2(3x-2))/(2(2x+1)^(1/2)))/(2x+1)

Factor out
(2x+1)^(-1/2) to get


((2x+1)^(-1/2)\left(3(2x+1)^(2/2)-(3x-2)\right))/(2x+1)=(6x+3-6x+2)/((2x+1)^(3/2))=(3x+5)/((2x+1)^(3/2))
User Josmith
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