In this case, we let u = 4x2x and then `y=sqrtu=u^(1/2)`.

Once again,

Using the chain rule, we firstly need to find:

`(dy)/(du)=1/2u^(-1/2)=1/(2sqrtu)=1/(2sqrt(4x^2-x)`

and

`(du)/(dx)=8x-1`

So

`(dy)/(dx)=(dy)/(du)(du)/(dx)`

`=1/(2sqrt(4x^2-x))(8x-1)`

`=(8x-1)/(2sqrt(4x^2-x))`

Note: How does `u^(-1//2)=1/sqrtu`? See:

Please support IntMath!