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

Answer 1

Isabella Ackerman

#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)#

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Answer 2

Christopher Agresta

To differentiate ( y = (x^2 - 4x)^3 ), apply the chain rule. First, find the derivative of the inner function ( u = x^2 - 4x ) with respect to ( x ), then multiply by the derivative of the outer function ( y = u^3 ) with respect to ( u ).

( u = x^2 - 4x )

( \frac{du}{dx} = 2x - 4 )

( y = u^3 )

( \frac{dy}{du} = 3u^2 )

Now, apply the chain rule:

( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} )

Substitute the derivatives:

( \frac{dy}{dx} = 3(x^2 - 4x)^2 \cdot (2x - 4) )

Therefore, ( \frac{dy}{dx} = 3(x^2 - 4x)^2(2x - 4) ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.