Question

1. Find the derivative with respect to x of x +1/x from first principle.
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Answer 1

If you mean
`f(x)=x+\frac1x`, then the derivative is

`\displaystyle f'(x) = \lim_(h\to0) \frac{\left(x+h+\frac1{x+h}\right) - \left(x+\frac1x\right)}h \\\\ = \lim_(h\to0) \frac{(x+h)-x}h + \lim_(h\to0) \frac{\frac1{x+h} - \frac1x}h \\\\ = \lim_(h\to0) \frac hh + \lim_(h\to0) (x-(x+h))/(hx(x+h)) \\\\ = \lim_(h\to0) 1 - \lim_(h\to0) \frac h{hx(x+h)} \\\\ = 1 - \lim_(h\to0) \frac1{x(x+h)} \\\\ = \boxed{1 - \frac1{x^2}}`

If you mean
`f(x) = \frac{x+1}x = 1 + \frac1x`, we know from above that

`\displaystyle \left(\frac1x\right)' = \lim_(h\to0) \frac{\frac1{x+h}-\frac1x}h = -\frac1{x^2}`

which leaves the constant term, whose derivative is

`\displaystyle (1)' = \lim_(h\to0)\frac{1 - 1}h = 0`

and so

`f'(x) = -\frac1{x^2}`