That is not going to help. It has nothing to do with this question. Asinus got confused.
You have
f(x)=x^2
And you know that
\( g'(x)=\frac{-1}{2x} \)
NOW can you integrate that to get the value of g(x)???
What is the
\(\int \frac{-1}{2x}\;dx\) ?
Normally, if you do not know the answer, you might do this :
\(\int \frac{-1}{2x}\;dx=\frac{-1}{2}\int x^{-1}\;dx = \frac{-1}{2}\int x^{-1+1}\;dx \)
But that leaves us with a power of 0 and that definitely is not right!
So this should remind you that the answer is a natural log !
Now can you or Asinus do this integral and get the function g ?
\(g(x)=\int \frac{-1}{2x}\;dx\)
.