# How do you differentiate g(x) =e^(1-x^2)sec using the product rule?

Aug 22, 2017

I assume this should be $g \left(x\right) = {e}^{1 - {x}^{2}} \sec x$

#### Explanation:

Note that

$\frac{d}{\mathrm{dx}} \left({e}^{1 - {x}^{2}}\right)$ may be found using thechain rule,

$\frac{d}{\mathrm{dx}} \left({e}^{1 - {x}^{2}}\right) = - 2 x {e}^{1 - {x}^{2}}$

And $\frac{d}{\mathrm{dx}} \left(\sec x\right) = \sec x \tan x$, so

$\frac{d}{\mathrm{dx}} \left({e}^{1 - {x}^{2}}\right) = - 2 x {e}^{1 - {x}^{2}} \sec x + {e}^{1 - {x}^{2}} \sec x \tan x$

Factor and rewrite to taste.