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

Answer 1

Christopher Agresta

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Answer 2

Aurora Alvarado

To differentiate ( y = (x^3y^{-2})^3 ) with respect to ( x ), we'll use the chain rule along with the power rule.

Let ( u = x^3y^{-2} ). Then ( y = u^3 ).

Using the power rule, we find ( \frac{dy}{du} = 3u^2 ).

Now, differentiate ( u ) with respect to ( x ), i.e., ( \frac{du}{dx} ):

[ \frac{du}{dx} = \frac{d}{dx}(x^3y^{-2}) ]

Applying the product rule:

[ \frac{du}{dx} = 3x^2y^{-2} - 2x^3y^{-3}\frac{dy}{dx} ]

Now, we can use the chain rule to find ( \frac{dy}{dx} ):

[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} ]

Substitute the derivatives we found:

[ \frac{dy}{dx} = (3u^2)(3x^2y^{-2} - 2x^3y^{-3}\frac{dy}{dx}) ]

Now, solve for ( \frac{dy}{dx} ).

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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.