What is the differentation of $1/x$ ?

Question

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

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

Which is correct?

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Answer 1 · 2017-07-30T12:44:41

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

$d/dx lnx = 1/x$

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

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 · 2017-07-30T12:49:31

Your are answer is correct

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

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