In is easy to apply the
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 - xWhen you take the derivatives of each of these, you get
1 1 u'(v) =The chain rule tells us to findor u'(w) =v w v'(x) = 1 w'(x) = -1
1 1 f'(x) =The second part asks for the derivative of-1 + x 1 - x
f(x) = ln(1 - x2)Again you'll need to use the
u(v) = ln(v) v(x) = 1 - x2Taking the derivatives you get
1 u'(v) = - v v'(x) = -2xThe chain rule tells you to find
-2x f'(x) =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 - x2
1 - x 1 + x -2x f'(x) =And since it is easy to show that-=(1 + x)(1 - x) (1 + x)(1 - x) (1 + x)(1 - x)
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