Karl's Calculus Tutor - Solution to Exercise 6.3-4

Solution to Exercise 6.3-4

The first part of this problem was to find the derivative of

   u(x)  =  e^-x²/2

Once again, this is a chain rule problem. You know this because this function is the composite, f(g(x)), of

   f(g)  =  e^g


            -x²
   g(x)  =     
             2

The chain rule says you have to find f'(g(x)) g'(x). This means you need to know the derivatives of f(x) and g(x). In this section you learned that e^x is its own derivative. You also know from earlier sections how to take the derivative of g(x). So you have

   f'(g)  =  e^g

   g'(x)  =  -x

Substituting those into the chain rule, you get

   u'(x)  =  -x e^-x²/2

which is the answer for the first derivative.

For the second derivative, u"(x), you will need to use the product rule. This is because you have to take the derivative of u'(x), and you have just seen that it is clearly a product. So if you let

   v(x)  =  -x

then (remembering from the original problem that u(x) = e^-x²/2) the problem is asking you to find the derivative of

  u'(x)  =  -x e^-x²/2  =  v(x) u(x)

The product rule tells you that this derivative will be v(x)u'(x) + v'(x)u(x). You have already found u'(x). And

   v'(x)  =  -1

So putting it all together you have

   u"(x)  =  (-x) (-x e^-x²/2)  +  (-1) e^-x²/2

which, by factoring out a e^-x²/2, is the same as

   u"(x)  =  (x² - 1) e^-x²/2

And that's your answer for the second derivative.

Recall that in this section we mentioned that if x is any real number, e^x 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 x = 0. If you put that value back into u(x), you get 1, so its maximum is at (0, 1). And you know it's a maximum and not a minimum because if you stick x = 0 into the second derivative, you get -1, which is negative (remember the rule from section 5).

Likewise, observe that the second derivative is x² - 1 times something that is never zero. So inflection points can occur only where x² - 1 = 0. And that occurs at x = ±1. Using a calculator, you can stick those x's back into u(x) and discover that the inflection points are at (±1, 0.60530659).

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