# What is the differentation of 1/x?

## If we write this as ${x}^{- 1}$, then it is $- \frac{1}{x} ^ 2$. But, my teacher has given it as $\log x$. Which is correct?

##### 2 Answers
Jul 30, 2017

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

$\frac{d}{\mathrm{dx}} \ln x = \frac{1}{x}$

#### Explanation:

$\frac{d}{\mathrm{dx}} \frac{1}{x} = \frac{d}{\mathrm{dx}} \left({x}^{- 1}\right) = - 1 \times {x}^{- 1 - 1} = - {x}^{- 2} = - \frac{1}{x} ^ 2$

You can easily derive it from first principles:

$\frac{d}{\mathrm{dx}} \frac{1}{x} = {\lim}_{h \to 0} \frac{1}{h} \left(\frac{1}{x + h} - \frac{1}{x}\right)$

$\frac{d}{\mathrm{dx}} \frac{1}{x} = {\lim}_{h \to 0} \frac{1}{h} \left(\frac{x - x - h}{x \left(x + h\right)}\right)$

$\frac{d}{\mathrm{dx}} \frac{1}{x} = {\lim}_{h \to 0} \frac{1}{h} \left(\frac{- h}{x \left(x + h\right)}\right)$

$\frac{d}{\mathrm{dx}} \frac{1}{x} = {\lim}_{h \to 0} - \frac{1}{x \left(x + h\right)}$

$\frac{d}{\mathrm{dx}} \frac{1}{x} = {\lim}_{h \to 0} - \frac{1}{x} ^ 2$

Jul 30, 2017

Your are answer is correct

#### Explanation:

The answer that your teacher had given to you is not differentiation, rather it is integration of $\frac{1}{x}$.

The method of integration exactly the converse of differentiation.
Example,
Differentiation of $\log x$ = $\frac{1}{x}$

Integration of $\frac{1}{x}$ = $\log x$

So, tell your teacher to make corrections

ENJOY MATHS !!!!!!!!