Question

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

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

Answer 1

$y=\dfrac{3x-2}{\sqrt{2x+1}}=\dfrac{3x-2}{(2x+1)^{1/2}}$

The quotient rule gives

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

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

$\dfrac{(2x+1)^{-1/2}\left(3(2x+1)^{2/2}-(3x-2)\right)}{2x+1}=\dfrac{6x+3-6x+2}{(2x+1)^{3/2}}=\dfrac{3x+5}{(2x+1)^{3/2}}$