# How do you find the derivative of d/( dx)(y = (3 x-2)/sqrt(2 x+1))?

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

Use the quotient rule that `(d(u/v))/dx =(v (du)/dx - v (dv)/dx)/v^2`

`y'=(sqrt(2x+1)(3)-(3x-2)(d(sqrt(2x+1)))/dx)/(2x+1)`

Now `(d(sqrt(2x+1)))/dx = 1/(2sqrt(2x+1))(2)=sqrt(2x+1)/(2x+1)` by the chain rule so

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

`y'=(((6x+3)-(3x-2))sqrt(2x+1))/(2x+1)^2=((3x+5)sqrt(2x+1))/(2x+1)^2` which is the answer.

We have to find the derivative of y = (3x-2)/sqrt(2 x+1))

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

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

use the product rule

y' = [(3x-2)]'*(2x + 1)^(-1/2) + (3x-2)*[(2x + 1)^(-1/2)]'

=> y' = 3*(2x + 1)^(-1/2) + (3x-2)*(-1/2)*2*(2x + 1)^(-3/2)

=> y' = 3*(2x + 1)^(-1/2) - (3x-2)*(2x + 1)^(-3/2)

**The required derivative is 3*(2x + 1)^(-1/2) - (3x-2)*(2x + 1)^(-3/2)**

Thank you so much!