Question

What is the differentation of `1/x`?

If we write this as `x^(-1)`, then it is `- 1/x^2`.

But, my teacher has given it as `logx`.

Which is correct?

Answer 1

I think you got confused, or your teacher did, as it is the other way around:

`d/dx lnx = 1/x`

Explanation:

`d/dx 1/x = d/dx (x^(-1)) = -1 xx x^(-1-1) = -x^(-2) = -1/x^2`

You can easily derive it from first principles:

`d/dx 1/x = lim_(h->0) 1/h(1/(x+h)-1/x)`

`d/dx 1/x = lim_(h->0) 1/h((x-x-h)/(x(x+h)))`

`d/dx 1/x = lim_(h->0) 1/h((-h)/(x(x+h)))`

`d/dx 1/x = lim_(h->0) -1/(x(x+h))`

`d/dx 1/x = lim_(h->0) -1/x^2`

Answer 2

Your are answer is correct

Explanation:

The answer that your teacher had given to you is not differentiation, rather it is integration of `1/x`.

The method of integration exactly the converse of differentiation.
Example,
Differentiation of `log x` = `1/x`

Integration of `1/x` = `log x`

So, tell your teacher to make corrections

ENJOY MATHS !!!!!!!!