# Derivative of an exponential

Is there a simplist f(x) such that

(d/dx)(exp(2[pi]i(f(x))))=(x/c+c/x)(exp(2[pi]i(f(x)))),

where c is a constant? My thanks in advance.

suffian
(d/dx) [ e2&pi; i f(x) ] = (x/c + c/x) e2&pi; i f(x) (the given)

set y = e2&pi; i f(x) (subtitution)

(d/dx) [y] = (x/c + c/x) y
(1/y) (dy/dx) = x/c + c/x
&int;(1/y)dy = &int;(x/c + c/x)dx
ln|y| = x2/(2c) + c ln|x| + k
eln|y| = ex2/(2c) + c ln|x| + k
|y| = K ex2/(2c) + c ln|x| (extract k)
e2&pi; i f(x) = K ex2/(2c) + c ln|x| (substitute original expression for y)

2&pi; i f(x) = x2/(2c) + c ln|x|
ln( K e2&pi; i f(x) ) = ln(K ex2/(2c) + c ln|x|)
f(x) = -ix2/(2&pi;+1c) - ic ln|x|/2&pi; + K
(same answer as hallsofivy except using 2&pi; instead of 2. note: i forgot to "un-e" the righthand side last time)
edit: made a serious typo plus changed to reflect that i stands for sqrt(-1)

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suffian, yes, "i" is here the imaginary number. How does that change your result?

HallsofIvy
Homework Helper
(d/dx)(exp(2i(f(x))))=(x/c+c/x)(exp(2i(f(x))))

(d/dx(exp(2if(x))))= 2i f'(x) exp(2if(x)) so this equation is the same as

2i f' (x)= x/c+ c/x which is a first order, non-linear differential equation for f. It is separable and can be written as

2i df= (x/c+ c/x)dx

Integrating both sides 2i f= (1/2c)x2+ c ln|x|+ D
( D is the constant of integration- an arbitrary complex number).

f(x)= (-i/4c)x2+ (-i/2)c ln|x|+ D

jeff