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?
Find the derivative of the function:
(x^(3)-8)/(x^(2)+9)
5 answers
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
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!
just curious -- if you forgot about the quotient rule, what did you try?
this web site dead kinda
1 answer