(i) Use part (ii) of Exercise 4-38 to show that, if f, g : (a, b) —»• Ж satisfy the equation f(t)g(t) = 1 for all t 6 (a, b) and are differentiable at x £ (a, b) then g'(x) = —f'(x)/f(x)2.
(ii) Explain why we can not deduce part (Hi) of Exercise 4-38 directly from part (i) of this exercise. Can we deduce the result of part (i) of this exercise directly from part (Hi) of Exercise 4-389 (Hi) Is the following statement true or false? If f, g : (a, b) —»• R are differentiable at x 6 (a, 6) and f(x)g(x) = 1 йегг д'(ж) = —f'(x)/f(x)2. Give a proof or counterexample.
Exercise 4.40. From time to time the eagle eyed reader will observe statements like
'f(x) —»• oo as x —»• —oo '
which have not been formally defined. If this really bothers her, she is probably reading the wrong book (or the right book but too early). It can be considered a standing exercise to fill in the required details.
In Appendix D, I sketch a method used in Beardon's elegant treatment [2] which avoids the need for such repeated definitions.