# exponential differentiation

• Jun 3rd 2007, 08:11 AM
confusedagain
exponential differentiation
f(x) = e^-x^2

f'(x) = e^-x^2 + (-2x)
f'(x) = -2xe^-x^2

f"(x) = -2x3^-x^2 +e^-x^2 + (-2x)
f"(x) =-4x(2e^-x^2)

how do you solve these types (e^-x^2) of problems to find the inflection point and concavity?
• Jun 3rd 2007, 08:42 AM
earboth
Quote:

Originally Posted by confusedagain
f(x) = e^-x^2

f'(x) = e^-x^2 + (-2x)
f'(x) = -2xe^-x^2

f"(x) = -2x3^-x^2 +e^-x^2 + (-2x)
f"(x) =-4x(2e^-x^2)

how do you solve these types (e^-x^2) of problems to find the inflection point and concavity?

Hello,

$f(x)=e^{-x^2}$ . Use chain rule to get the first derivation:

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

f'(x) = 0 if x = 0

Use product rule to get the second derivation:

$f''(x) = e^{-x^2} \cdot (-2) + (-2x) \cdot e^{-x^2} \cdot (-2x) = e^{-x^2} ( 4x^2 -2)$

f''(x) = 0 if 4x² - 2 = 0 that means $x=\sqrt{\frac{1}{2}} \ \vee \ x= -\sqrt{\frac{1}{2}}$
• Jun 3rd 2007, 09:46 AM
curvature
the curv
the figure show the inflection points
• Jun 3rd 2007, 10:19 AM
Jhevon
Quote:

Originally Posted by curvature
the figure show the inflection points

umm...there's no figure