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How do you find the derivative of #(x^3-3x^2+4)/x^2#?

Answer 1

Cameron Alexander

Using the quotient rule, which states that

Be #y=f(x)/g(x)#, #(dy)/(dx)=(f'(x)g(x)-f(x)g'(x))/(g(x))^2#

Applying the rule:

#(dy)/(dx)=((3x^2-6x)x^cancel(2)-(x^3-3x^2+4)(2cancel(x)))/x^(cancel(4)3)#

#(dy)/(dx)=(3x^3cancel(-6x^2)-2x^3cancel(+6x^2)-8)/x^3=(x^3+8)/x^3#

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Answer 2

Adam Adkins

To find the derivative of (\frac{{x^3 - 3x^2 + 4}}{{x^2}}), you can use the quotient rule. The quotient rule states that if you have a function in the form (\frac{{f(x)}}{{g(x)}}), then the derivative is given by (\frac{{f'(x)g(x) - f(x)g'(x)}}{{[g(x)]^2}}).

Let (f(x) = x^3 - 3x^2 + 4) and (g(x) = x^2). Then, (f'(x) = 3x^2 - 6x) and (g'(x) = 2x).

Now, applying the quotient rule, we get:

[ \frac{{d}}{{dx}}\left(\frac{{x^3 - 3x^2 + 4}}{{x^2}}\right) = \frac{{(3x^2 - 6x)(x^2) - (x^3 - 3x^2 + 4)(2x)}}{{(x^2)^2}} ]

Simplify this expression to find the derivative.

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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.