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| rev | Admin | (Created page with "'''Solution: B''' <math>\delta=\ln (1+i)</math> and <math>d=i /(1+i)=1-1 /(1+i)</math>. Thus <math display = "block"> \begin{aligned} \frac{d f}{d i}=\frac{d}{d i} \frac{(1+i) \ln (1+i)}{i} &=\frac{d}{d i}(1+1 / i) \ln (1+i) = \frac{-1}{i^2} \ln (1+i)+\left(1+\frac{1}{i}\right) \frac{1}{1+i} \\ &=\frac{-\ln (1+i)}{i^2}+\frac{1}{i}=\frac{-\delta}{i^2}+\frac{1}{i}=\frac{i-\delta}{i^2} \end{aligned} </math> '''References''' {{cite web |url=https://web2.uwindsor.ca/math/...") | Nov 26'23 at 17:51 | +652 |