anonymous one year ago Can someone help me find the derivative of the problem below? I'll give a medal! :)

1. anonymous

$\frac{ 5 }{ (2x)^3 }$

do you want to use quotient rule or product rule? personally I think product rule is easier xD

3. anonymous

Whatever you think is easier, I'm okay with.

mmmkk well do you know product rule? :) $\frac{ d }{ dx } [f(x) * g(x)] ~~=~~ f(x)*g'(x) + g(x)*f'(x)$right? :)

5. anonymous

Yes, I am familiar with it

mmm I'm not awake we can rewrite the given equation as $5*(2x)^{-3}$right?

7. anonymous

Yes

turns out we don't need product rule xD (cuz the numeator is a constant...) ingore what me said before we use power rule and chain rule $5*(2x)^{-3} \rightarrow 5(-3)*(2x)^{-4}*(\frac{ d }{ dx }2x)$

tell me if you get confused xD

10. anonymous

11. anonymous

You'd get -30(-2x)^4

12. anonymous

-30(2x)^4 sorry the 2s not negative

mhmm then we simplify and get $-15 * (2x)^{-4} * (2) \rightarrow -30 * (2x)^{-4} \rightarrow \frac{ -30 }{ (2x)^{4} }$$$\huge \rightarrow \frac{-30}{16x^4} \rightarrow \frac{-15}{8x^4}$$

14. freckles

you can also just use power rule... if you write $5(2x)^{-3} =5(2)^{-3}x^{-3} =\frac{5}{2^3}x^{-3}=\frac{5}{8}x^{-3} \\ \text{ then differentiate using power rule }$

yay math thanks @freckles :)

16. anonymous

Thank you, both of you. :)