Question

Find the derivative using the quotient rule:

1 - (64/x^3)

Answer 1

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 \(g(x)=x^3\).

Answer 2

Remember that derivative of constant is zero.

Answer 3

Denominator is \(x^3\) or \(x^2\)?

Answer 4

These are the steps I did

d/dx (1) = 0

\[\frac{ d }{ dx } - \frac{ 64 }{ x ^{3} } =\]

Answer 5

\[\frac{ - x ^{3}(0) - (64)3x ^{2} }{ (x ^{3})^{2} }\]

Answer 6

64 can be written as 4^3

Answer 7

@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.

Answer 8

\[\frac{ -0 -192x ^{2}}{ x ^{6} } = - \frac{ 192 }{ x ^{4} }\]

Answer 9

Supposed to be \[\frac{ \text- x ^{3}(0) - (\color{red}{\textbf-}64)3x ^{2} }{ (x ^{3})^{2} }\]

Answer 10

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

\[64(-3x ^{-3-1}) = -192x ^{-4} = - \frac{ 192 }{ x ^{4} }\]

Answer 11

@geerky42 Thank you for your help

Answer 12

Yeah power rule is preferable, but you were asked to use quotient rule lol.

Careful with minus sign. Final answer should be \(\dfrac{192}{x^4}\).

Answer 13

@geerky 42 You are correct, the final answer should not have the negative sign.