# How do you differentiate f(x)=e^(x^2)/(e^x-x^2) using the quotient rule?

Oct 7, 2016

$\setminus \frac{\left({e}^{{x}^{2}} \cdot 2 x\right) \left({e}^{x} - {x}^{2}\right) - {e}^{{x}^{2}} \left({e}^{x} - 2 x\right)}{{\left({e}^{x} - {x}^{2}\right)}^{2}}$

#### Explanation:

$\frac{d}{\mathrm{dx}} f \frac{x}{g} \left(x\right) = \setminus \frac{f ' g - f g '}{{g}^{2}}$

Let's compute all the functions we need:

• $f \left(x\right) = {e}^{{x}^{2}}$
• $f ' \left(x\right) = {e}^{{x}^{2}} \cdot 2 x$
• $g \left(x\right) = {e}^{x} - {x}^{2}$
• $g ' \left(x\right) = {e}^{x} - 2 x$
• ${g}^{2} \left(x\right) = {\left({e}^{x} - {x}^{2}\right)}^{2}$

Let's plug these values into the formula:

\frac{f'g-fg'}{g^2} = \frac{(e^{x^2}*2x)(e^x-x^2)-e^{x^2}(e^x-2x)}{(e^x-x^2)^2}