Question

How do you find the derivative of `y=(2/(x-1)-x^-3)^4`?

Answer 1

`y'=4(2/(x-1) - x^-3)^3 ((-2)/(x-1)^2+3x^-4)`

Explanation:

Use the quotient, power, and chain rules.

`y = (2/(x-1) - x^-3)^4`

Let `u=2/(x-1)` and so through the quotient rule `u'=((x-1)*d/dx2 - 2d/dx(x-1))/(x-1)^2` which simplifies to `u'=(-2)/(x-1)^2`

also `v=x^-3` and so `v'=-3x^-4`

`y=(u-v)^4` and so we get `y'=4(u-v)^3(u'-v')` through the chain rule.

substituting we get;

`y'=4(2/(x-1) - x^-3)^3 ((-2)/(x-1)^2+3x^-4)`