Answer
See explanations.
Work Step by Step
Step 1. Given $f(x)=\frac{x^2}{x^2+1}=\frac{x^2+1-1}{x^2+1}=1-\frac{1}{x^2+1}$, we have $f'(x)=\frac{2x}{(x^2+1)^2}$
Step 2. Given $g(x)=-\frac{1}{x^2+1}$, we have $g'(x)=\frac{2x}{(x^2+1)^2}$
Step 3. Since both functions have the same derivative, there is a constant ($C=1$ in this case) such that $f(x)=g(x)+C$ and their graphs should be the same shape with a vertical shift.
