If we let *u *= 2*x*^{3} - 1 then *y *= *
u*^{4}.

So now

*y*is written as a power of*u*; and*u*is a function of*x*[*u*=*f*(*x*) ].

To find the derivative of such an expression, we can use our new rule:

`d/(dx)u^n=n u^(n-1)(du)/(dx`

where *u *= 2*x*^{3} − 1 and *n* =
4.

So

`(dy)/(dx)=n u^(n-1)(du)/(dx)`

`=[4(2x^3-1)^3][6x^2]`

`=24x^2(2x^3-1)^3`

We could, of course, use the **chain rule**, as
before**:**

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