# Thread: power derivative

1. ## power derivative

Hi,

I was wondering how I would convert $\displaystyle \frac{1}{3x^3}$ and $\displaystyle \frac{1}{3x}$ to "power form". For example $\displaystyle \frac{2}{x^3}$ would be $\displaystyle 2x^-3$ but when I try the same logic with the first two, $\displaystyle \frac{1}{3x^3}$ would become $\displaystyle {3x^-3}$ but that's wrong? When 1 is the numberator I seem to get them wrong.

2. Originally Posted by Ifailatmaths
Hi,

I was wondering how I would convert $\displaystyle \frac{1}{3x^3}$ and $\displaystyle \frac{1}{3x}$ to "power form". For example $\displaystyle \frac{2}{x^3}$ would be $\displaystyle 2x^-3$ but when I try the same logic with the first two, $\displaystyle \frac{1}{3x^3}$ would become $\displaystyle {3x^-3}$ but that's wrong? When 1 is the numberator I seem to get them wrong.
The constants stay where they are in the fraction.

$\displaystyle \frac{1}{3x^2} = \frac{1}{3} \times \frac{1}{x^2}$

The law of indices states that $\displaystyle \frac{1}{a^b} = a^{-b}$

Hence we get:

$\displaystyle \frac{1}{3} \times \frac{1}{x^2} = \frac{1}{3} \times x^{-2} = \frac{x^{-2}}{3}$

3. Thanks mush but Im trying to get it into the form so I can differentiate and for that don't I need it without any fractions and just powers?

4. Originally Posted by Ifailatmaths
Thanks mush but Im trying to get it into the form so I can differentiate and for that don't I need it without any fractions and just powers?
Nope. When you differentiate a term $\displaystyle \frac{p}{q} x^{n}$ where p and q are constants, then the constants come OUTSIDE the integration.

$\displaystyle \frac{d}{dx} (\frac{x^{-2}}{3})$

$\displaystyle =\frac{d}{dx} (\frac{1}{3} \times x^{-2})$

$\displaystyle =\frac{1}{3} \times \frac{d}{dx} (x^{-2})$

$\displaystyle =\frac{1}{3} \times (-2x^{-3})$

$\displaystyle =\frac{-2x^{-3}}{3}$

$\displaystyle =\frac{-2}{3x^{3}}$