# Chapter 3 - Differentiation - 3.3 Product and Quotient Rules - Exercises - Page 121: 12

$$f'(x) = \frac{e^x(x^2-2x+1)}{(x^2+1)^2}$$

#### 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: $$f'(x) = \frac{e^x(x^2-2x+1)}{(x^2+1)^2}$$

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