Find the derivative of the function:

(x^(3)-8)/(x^(2)+9)

just curious -- if you forgot about the quotient rule, what did you try?

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To find the derivative of the function (x^3 - 8)/(x^2 + 9), we can apply the quotient rule. The quotient rule states that if you have a function f(x) = g(x)/h(x), then the derivative of f(x) is given by:

f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2.

Let's apply the quotient rule to the given function.

First, we need to find the derivative of the numerator and the denominator separately.

1. For the numerator:
Using the power rule, the derivative of x^3 is 3x^(3-1) = 3x^2.
The derivative of the constant term -8 is zero since it does not depend on x.

So, the derivative of the numerator, g'(x), is 3x^2.

2. For the denominator:
Using the power rule, the derivative of x^2 is 2x.
The derivative of the constant term 9 is zero since it does not depend on x.

So, the derivative of the denominator, h'(x), is 2x.

Now, we substitute these derivatives into the quotient rule formula:
f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2
= (3x^2 * (x^2 + 9) - (x^3 - 8) * 2x) / (x^2 + 9)^2
= (3x^4 + 27x^2 - 2x^4 + 16x) / (x^2 + 9)^2
= (x^4 + 27x^2 + 16x) / (x^2 + 9)^2.

Therefore, the derivative of the function (x^3 - 8)/(x^2 + 9) is (x^4 + 27x^2 + 16x) / (x^2 + 9)^2.

The rule is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator AND then put this answer over the denominator squared.

The deriv. of the numerator is 3x^2 and the deriv of the denom. is 2x

Can you finish from here?

y = (x^3-8)/(x^2+9)

If y = u/v, y' = (vu' - uv')/v^2, so

y = ((3x^2)(x^2+9) - (x^3-8)(2x))/(x^2+9)^2

= (3x^4 + 27x^2 - 2x^4 + 16x)/(x^2+9)^2

= (x^4 + 27x^2 + 16x)/(x^2+9)^2

Yeah, I was forgetful and forgot it was the Quotient Rule. Thank you all, I got it!