# [SOLVED] Help with domain

• Oct 22nd 2009, 05:50 PM
el123
[SOLVED] Help with domain
$\displaystyle f(x)=e^{-x^2}$

Find domain.
Find first and second derivative

Can some one please take me through this?
• Oct 22nd 2009, 06:09 PM
skeeter
Quote:

Originally Posted by el123
$\displaystyle f(x)=e^{-x^2}$

Find domain.
Find first and second derivative

Can some one please take me through this?

domain is all real numbers

$\displaystyle f'(x) = e^{-x^2} \cdot (-2x)$

$\displaystyle f''(x) = e^{-x^2} \cdot (-2) + (-2x) \cdot e^{-x^2} \cdot (-2x)$

you can clean up the algebra
• Oct 22nd 2009, 06:14 PM
pickslides
Quote:

Originally Posted by el123
$\displaystyle f(x)=e^{-x^2}$

Find domain.

The function is defined for all values of x as you can square any real number and raise any real number over e.

Quote:

Originally Posted by el123
$\displaystyle f(x)=e^{-x^2}$

Find first and second derivative

Consider

If $\displaystyle y = e^{f(x)} \Rightarrow \frac{dy}{dx} = f'(x)e^{f(x)}$

to find the 1st.

Repeat this in conjunction with the product rule on the 1st to obtain the second.
• Oct 22nd 2009, 06:58 PM
ux0
Quote:

Originally Posted by el123
$\displaystyle f(x)=e^{-x^2}$

Find domain.
Find first and second derivative

Can some one please take me through this?

The easiest way to understand domain is to ask yourself what values of X can i put into the function f(x)?

The derivative is just

$\displaystyle e^udu$

where your

$\displaystyle u=-X^2$

$\displaystyle f'(x)= -2xe^{-X^2}$

and ill leave you to find the second one, in which case you will have to use the chain rule...

$\displaystyle f(x)g(x)=f'(x)g(x)+f(x)g'(x)$