How do you find the derivative of #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)#