Answer
$$f'(x) = \frac{e^x(x^2-2x+1)}{(x^2+1)^2}$$
Work Step by Step
We know by the quotient rule that: $$(\frac{f}{g})' = \frac{gf'-fg'}{g^2}$$
Let $$f = e^x,\ g = x^2+1$$
We can calculate:
$$f' = e^x,\ and\ g' = 2x$$
Now, by plugging everything back into the equation from above, we get:
$$f'(x) = \frac{(x^2+1)e^x - 2x(e^x)}{(x^2+1)^2}$$
Now, we factor out the e and get:
$$f'(x) = \frac{e^x(x^2-2x+1)}{(x^2+1)^2}$$
