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

3 Answers
Apr 9, 2018

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#

Apr 9, 2018

#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"#

Apr 9, 2018

#-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#