# What is the function of the derivative f'(x)=1/x^2?

Apr 9, 2018

To find the function you need to integrate the derivative.

#### Explanation:

This means we have to evaluate:

$\int$ $\frac{1}{x} ^ 2$ $\mathrm{dx}$

$= \int$ ${x}^{-} 2$ $\mathrm{dx}$

$= - {x}^{-} 1 + C$

Therefore

$f \left(x\right) = C - \frac{1}{x}$

Apr 9, 2018

$f \left(x\right) = - \frac{1}{x} + c$

#### Explanation:

$\text{to find the function given the derivative we integrate}$

$\Rightarrow f \left(x\right) = \int \frac{1}{x} ^ 2 \mathrm{dx} = \int {x}^{- 2} \mathrm{dx}$

$\text{integrate using the "color(blue)"power rule}$

int(ax^n)=a/(n+1)x^(n+1);n!=-1

$\Rightarrow \int {x}^{- 2} \mathrm{dx} = - 1 {x}^{\left(- 2 + 1\right)} = - 1 {x}^{- 1} = - \frac{1}{x} + c$

$\text{where c is the constant of integration}$

Apr 9, 2018

$- \frac{1}{x} + C$

#### Explanation:

Given: $f ' \left(x\right) = \frac{1}{x} ^ 2$

Then, $f \left(x\right) = \int f ' \left(x\right) \setminus \mathrm{dx}$

$\implies f \left(x\right) = \int \frac{1}{x} ^ 2 \setminus \mathrm{dx}$

$= \int {x}^{-} 2 \setminus \mathrm{dx}$

Use the power rule, which states that, $\int {a}^{n} \setminus \mathrm{dx} = \frac{{a}^{n + 1}}{n + 1}$.

$= \frac{{x}^{- 2 + 1}}{- 2 + 1}$

$= \frac{{x}^{-} 1}{-} 1$

$= - {x}^{-} 1$

$= - \frac{1}{x}$

Now, we need to add a constant, since this is not a definite integral.

$= - \frac{1}{x} + C$