201 Complex Numbersfor a very smallt. Since ∆xapproaches zero as a limit, for any fixedx,∆xxcan certainly be made as small as we wish. Therefore, we can set∆xx=t, andconclude∆xx= ln1 +∆xx.Thus,d(lnx)dx=lim∆x→01∆xln1 +∆xx=lim∆x→01∆x∆xx=1x.This in turn meansd (lnx) =dxxordxx= lnx+c,(1.12)wherecis the constant of integration. It is well known that because ofdxn+1dx= (n+ 1)xn,we havexndx=xn+1(n+ 1)+c.This formula holds for all values ofnexcept forn=−1, since then thedenominatorn+ 1 is zero. This had been a difficult problem, but now we seethat (1.12) provides the “missing case.”In numerous phenomena, ranging from population growth to the decay ofradioactive material, in which the rate of change of some quantity is propor-tional to the quantity itself. Such phenomenon is governed by the differentialequationdydt=ky,wherekis a constant that is positive ifyis increasing and negative ifyisdecreasing. To solve this equation, we write it asdyy=kdtand then integrate both sides to getlny=kt+c,ory= ekt+c= ektec.Ify0denotes the value ofywhent= 0, theny0= ecandy=y0ekt.This equation is called the law of exponential change.