How do you differentiate $y= 3 / (sqrt(2x+1)$ ?

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Answer 1 · 2016-11-01T16:36:08

$dy/dx=-3/(2x+1)^(3/2)$

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

$dy/dx=-3/2 (2x+1)^(-3/2) *2$

$dy/dx=-3/(2x+1)^(3/2)$

How do you differentiate y= 3 / (sqrt(2x+1)y= 3 / (sqrt(2x+1)?