## anonymous one year ago Find the derivative using the quotient rule: 1 - (64/x^3)

1. geerky42

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

2. geerky42

Remember that derivative of constant is zero.

3. geerky42

Denominator is $$x^3$$ or $$x^2$$?

4. anonymous

These are the steps I did d/dx (1) = 0 $\frac{ d }{ dx } - \frac{ 64 }{ x ^{3} } =$

5. anonymous

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

6. triciaal

64 can be written as 4^3

7. anonymous

@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.

8. anonymous

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

9. geerky42

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

10. anonymous

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

11. anonymous

@geerky42 Thank you for your help

12. geerky42

Yeah power rule is preferable, but you were asked to use quotient rule lol. Careful with minus sign. Final answer should be $$\dfrac{192}{x^4}$$.

13. anonymous

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

14. triciaal

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