Optimization Part II

[MarkFL]

This may be way too simplistic, but

[MATH]a > 0 \implies a^{-b} \equiv \dfrac{1}{a^b}.[/MATH]
Therefore,

[MATH]x > 0 \text { and } f(x) = \dfrac{1}{x^n} = x^{-n} \implies f'(x) = x^{\{(-n)-1)\}} = x^{\{-(n+1)\}} = \dfrac{1}{x^{(n+1)}}.[/MATH]

Don't forget to "bring down" your exponent before subtracting 1 from it.

 

[Steven G]

This may be way too simplistic, but

[MATH]a > 0 \implies a^{-b} \equiv \dfrac{1}{a^b}.[/MATH]
Therefore,

[MATH]x > 0 \text { and } f(x) = \dfrac{1}{x^n} = x^{-n} \implies f'(x) = x^{\{(-n)-1)\}} = x^{\{-(n+1)\}} = \dfrac{1}{x^{(n+1)}}.[/MATH]

You made your point quite well but there is an error in the derivative which you can't fix anymore because of the edit time limit we have. In any case, for the OP we need to mention that [MATH] f'(x) = -nx^{\{(-n)-1)\}} = -nx^{\{-(n+1)\}} = \dfrac{-n}{x^{(n+1)}}.[/MATH]

 

[Harry_the_cat]

You made your point quite well but there is an error in the derivative which you can't fix anymore because of the edit time limit we have. In any case, for the OP we need to mention that [MATH] f'(x) = nx^{\{(-n)-1)\}} = nx^{\{-(n+1)\}} = \dfrac{n}{x^{(n+1)}}.[/MATH]

There should be a negative sign in front!

 

[Steven G]

There should be a negative sign in front!

Just edited my post. Thanks!!

 

[Hckyplayer8]

Thank you all!