How do you differentiate $y = {(x^{2} - 4 x)}^{3}$ $y=(x^2-4x)^3$ ?

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Answer 1 · 2016-11-16T22:31:11

$y'=6(x^2-4x)^2(x-2)$

Use the chain rule, which is basically "derivative of the outside times derivative of the inside":

$y=(x^2-4x)^3$

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

Factor a 2 out of the last term to simplify:
$y'=3(x^2-4x)^2(2)(x-2)$

$y'=6(x^2-4x)^2(x-2)$

How do you differentiate y=(x^2-4x)^3y=(x2−4x)3y=(x^2-4x)^3?