Solution to Exercise 6.3-2

In is easy to apply the chain rule to each of the summands in

   f(x)  =  ln(1 + x) + ln(1 - x)

Each summand is a composite, u(v(x)) and u(w(x)) respectively, where

   u(v)  =  ln(v)       or      u(w)  =  ln(w)

   v(x)  =  1 + x

   w(x)  =  1 - x

When you take the derivatives of each of these, you get

             1                        1
   u'(v)  =         or      u'(w)  =   
             v                        w

   v'(x)  =  1

   w'(x)  =  -1

The chain rule tells us to find u'(v(x)) v'(x) and u'(w(x)) w'(x) and then sum them together. By substitution of the expressions above you get

f'(x) =

1 - 1 + x

1 1 - x

The second part asks for the derivative of

   f(x)  =  ln(1 - x²)

Again you'll need to use the chain rule. This time you have

   u(v)  =  ln(v)

   v(x)  =  1 - x²

Taking the derivatives you get

             1
   u'(v)  =  -
             v

   v'(x)  =  -2x

The chain rule tells you to find u'(v(x)) v'(x). Substituting the expressions above we get.

              -2x
   f'(x)  =        
             1 - x²

The problem asks you to show that this derivative is equal to the the first one you did. At first they seem different. But with a little algebra you can turn the first expression into the second. Simply put the two summands that constitute that first expression over a common denominator. That common denominator would be the product of the two denominators, (1 + x)(1 - x). So the first expression becomes

f'(x) =

1 - x - (1 + x)(1 - x)

1 + x = (1 + x)(1 - x)

-2x (1 + x)(1 - x)

And since it is easy to show that (1 + x)(1 - x) = 1 - x², the identity is proved.

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