Question

Is the derivative of f(x)=x+ 1/x
f'(x)=1-x^-2 ?

Answer 1

yes, although if you want a useful f0rm you would write
\[f'(x)=1-\frac{1}{x^2}\]

Answer 2

ok thanks ^_^

Answer 3

i'm trying to find the critical numbers

Answer 4

wait, maybe it is\[\frac{x+1}{x}\]

Answer 5

well there you go then a perfectly good example of what i just said
now it is easy to find the critical numbers
oh yeah, maybe it is

Answer 6

but the space after the first x, indicate the correctness of your interpretation, satelline.

Answer 7

x is undefined when the denominator equals zero rihgt?

Answer 8

actually i am going to stick with my first answer
\[f'(x)=\frac{x^2-1}{x^2}\]

Answer 9

yes, but i would not really call that a critical number
your original function is not defined at 0 so it would be a miracle of the derivative was

Answer 10

The denominator tells you where you have a vertical asymptote, I believe.

Answer 11

Or yeah, where the function is undefined, because there is a VA there.

Answer 12

yes.

Answer 13

So solve: \(x^2-1=0\)

This wll give you your critical points.

Answer 14

\[\pm1\]

Answer 15

yes.

Answer 16

yup

Answer 17

again, that is why you want to write the derivative of \(\frac{1}{x}\) as \(-\frac{1}{x^2}\) rather than \(x^{-2}\) which is correct but does not help your cause at all

Answer 18

similarly, the derivative of \(\sqrt{x}\) is \(\frac{1}{2\sqrt{x}}\) not \(\frac{1}{2}x^{-\frac{1}{2}}\)