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How do you integrate #y=(2x^3)/(4-x)# using the quotient rule?

Answer 1

William Abdalla

#(4 x^(2) (6 - x)) / ((4 - x)^(2))#

The question should be asking how to differentiate the given function, and not how to integrate it.

We have: #y = (2 x^(3)) / (4 - x)#

This function can be differentiated using the "quotient rule":

#=> y' = ((4 - x) (6 x^(2)) - (2 x^(3)) (- 1)) / ((4 - x)^(2))#

#=> y' = (24 x^(2) - 6 x^(3) + 2 x^(3)) / ((4 - x)^(2))#

#=> y' = (24 x^(2) - 4 x^(3)) / ((4 - x)^(2))#

#=> y' = (4 x^(2) (6 - x)) / ((4 - x)^(2))#

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Answer 2

Cameron Alexander

To integrate ( y = \frac{2x^3}{4-x} ) using the quotient rule, follow these steps:

Rewrite the expression: ( y = 2x^3 \cdot (4-x)^{-1} )
Apply the quotient rule: ( \int \frac{f'(x)}{g(x)} ,dx = \ln|g(x)| + C )
Let ( f(x) = 2x^3 ) and ( g(x) = 4-x )
Find ( f'(x) ) and ( g'(x) )
- ( f'(x) = 6x^2 )
- ( g'(x) = -1 )
Substitute into the quotient rule formula:
- ( \int \frac{2x^3}{4-x} ,dx = \ln|4-x| + C )

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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.