The first part of this problem was to find the derivative of
u(x) = e-x2/2Once again, this is a
f(g) = eg -x2 g(x) =The chain rule says you have to find2
f'(g) = eg g'(x) = -xSubstituting those into the chain rule, you get
u'(x) = -x e-x2/2which is the answer for the first derivative.
For the second derivative, u"(x), you will need to use the
v(x) = -xthen (remembering from the original problem that
u'(x) = -x e-x2/2 = v(x) u(x)The product rule tells you that this derivative will be
v'(x) = -1So putting it all together you have
u"(x) = (-x) (-x e-x2/2) + (-1) e-x2/2which, by factoring out a e-x2/2, is the same as
u"(x) = (x2 - 1) e-x2/2And that's your answer for the second derivative.
Recall that in this section we mentioned that if x is any real number, ex is always positive. Hence e-x2/2 must also be positive, because -x2/2 is always real. This fact is useful in finding the maximum and the inflection points.
To find the maximum, observe that the derivative of u(x) is
the product of -x with something that is never zero.
So the only places where u'(x) can possibly be zero are
where -x is zero, and that only happens at
Likewise, observe that the second derivative is
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