Question

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

Answer 1

To find the function you need to integrate the derivative.

Explanation:

This means we have to evaluate:

`int` `1/x^2` `dx`

`= int` `x^-2` `dx`

`= - x^-1 + C`

Therefore

`f(x) = C - 1/x`

Answer 2

`f(x)=-1/x+c`

Explanation:

`"to find the function given the derivative we integrate"`

`rArrf(x)=int1/x^2dx=intx^(-2)dx`

`"integrate using the "color(blue)"power rule"`

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

`rArrintx^(-2)dx=-1x^((-2+1))=-1x^(-1)=-1/x+c`

`"where c is the constant of integration"`

Answer 3

`-1/x+C`

Explanation:

Given: `f'(x)=1/x^2`

Then, `f(x)=intf'(x) \ dx`

`=>f(x)=int1/x^2 \ dx`

`=intx^-2 \ dx`

Use the power rule, which states that, `inta^n \ dx=(a^(n+1))/(n+1)`.

`=(x^(-2+1))/(-2+1)`

`=(x^-1)/-1`

`=-x^-1`

`=-1/x`

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

`=-1/x+C`