Finding dy/dx of y=4-1/x

[The riddler]

1. Hello everyone, my problem is with differentation. It seems to be quite simple but i can't understand it. Its basically find dy/dx of this equation:
y=4-1/x



2. The only relevant equation is dy/dx



3. I know how to differentate the equation to get y'=4-x^-1 but i don't know how to get from there to dy/dx. To be honest I am not really sure what dy/dx is. I have been told the answer is dy/dx=x^-2. Can someone please give me a clear explanation to what dy/dx means and explain to me how i get from the y' equation to dy/dx. Thanks for any replies :)

 

[madah12]

first y'(x) is dy/dx, second y = 4 - x^-1 to get to y' ie dy/dx you have to differentiate it but the only thing you did as I seen is rewrote y from 4-1/x to 4- x^-1 that is not differentiation.

 

[The riddler]

Don't worry i see where i was going wrong i understand now for some reason i thought Differentiate ment somthing else, thanks for replying anywayz