Differentiation help :)

I've been given the question:

Find the stationary points of $y=\frac{1}{x} + \frac{1}{x^2} - \frac{1}{x^3}$ and determine the nature of each.

I think I've been approaching it in the wrong way because I'm completely stuck for an answer. Would anyone be able to talk me through how to solve it? (I'm on OCR AS Level if it helps)

Thanks
xxx

I've been given the question:

Find the stationary points of $y=\frac{1}{x} + \frac{1}{x^2} - \frac{1}{x^3}$ and determine the nature of each.

I think I've been approaching it in the wrong way because I'm completely stuck for an answer. Would anyone be able to talk me through how to solve it? (I'm on OCR AS Level if it helps)

Thanks
xxx

(Remember that $\frac{d}{dx}\left(\frac{1}{x}\right)=\mathrm{ln}(x)$. )

First, differentiate. Stationary points will be at the zeroes of the derivative.

(Remember that $\frac{d}{dx}\left(\frac{1}{x}\right)=\mathrm{ln}(x)$. )

What exactly are you stuck on?

I've been given the question:

Find the stationary points of $y=\frac{1}{x} + \frac{1}{x^2} - \frac{1}{x^3}$ and determine the nature of each.

I think I've been approaching it in the wrong way because I'm completely stuck for an answer. Would anyone be able to talk me through how to solve it? (I'm on OCR AS Level if it helps)

Thanks
xxx

oh really..

I changed the fractions to indices
$x^{-1}+x^{-2}-x^{-3}$
(which I cant say made it much easier and differentiated it to
$\frac{dy}{dx}=-x^{-2}-2x^{-3}+3x^{-4}$
and set $\frac{dy}{dx}$ to 0.

Im not really sure where to go from there though. I did try factorising it so x=0 at one point but I really am completely stuck.

Looks like I need some differentiation help. All fixed.

I've been given the question:

Find the stationary points of $y=\frac{1}{x} + \frac{1}{x^2} - \frac{1}{x^3}$ and determine the nature of each.