How do you find the derivative of $y = {(\frac{2}{x - 1} - x^{- 3})}^{4}$ $y=(2/(x-1)-x^-3)^4$ ?

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Answer 1 · 2017-08-27T14:28:03

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

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;

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

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