Find the derivative using the quotient rule: 1 - (64/x^3)

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Find the derivative using the quotient rule: 1 - (64/x^3)

Mathematics
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Really, quotient rule? Well, this rule states that \(\left(\dfrac{f(x)}{g(x)}\right)' = \dfrac{f'(x)~g(x) - f(x)~g'(x)}{(~g(x)~)^2}\) Here, you have \(f(x)=-64\) and \(g(x)=x^3\).
Remember that derivative of constant is zero.
Denominator is \(x^3\) or \(x^2\)?

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These are the steps I did d/dx (1) = 0 \[\frac{ d }{ dx } - \frac{ 64 }{ x ^{3} } =\]
\[\frac{ - x ^{3}(0) - (64)3x ^{2} }{ (x ^{3})^{2} }\]
64 can be written as 4^3
@geerky42 denominator is x^3. Originally I made a mistake in solving the problem; as I'm typing my steps I think I realized my mistake.
\[\frac{ -0 -192x ^{2}}{ x ^{6} } = - \frac{ 192 }{ x ^{4} }\]
Supposed to be \[\frac{ \text- x ^{3}(0) - (\color{red}{\textbf-}64)3x ^{2} }{ (x ^{3})^{2} }\]
I do think though it would be easier to solve by rewriting it as \[64\frac{ d }{ dx }\frac{ 1 }{ x ^{3} } =64\frac{ d }{ dx }x ^{-3} = \] and using the power rule \[64(-3x ^{-3-1}) = -192x ^{-4} = - \frac{ 192 }{ x ^{4} }\]
@geerky42 Thank you for your help
Yeah power rule is preferable, but you were asked to use quotient rule lol. Careful with minus sign. Final answer should be \(\dfrac{192}{x^4}\).
@geerky 42 You are correct, the final answer should not have the negative sign.
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