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| rev | Admin | (Created page with "'''Solution A''' <math display="block"> \begin{aligned} & \frac{d}{d v} \delta=\frac{d}{d v} \ln (1+i)=\frac{d}{d v} \ln \left(v^{-1}\right)=\frac{d}{d v}(-\ln (v))=\frac{-1}{v} . \\ & \frac{d}{d i} d=\frac{d}{d i} \frac{i}{1+i}=\frac{1}{(1+i)^2} \end{aligned} </math> Hence <math display = "block">\left(\frac{d}{d v} \delta\right)\left(\frac{d}{d i} d\right)=\frac{-1}{v} \frac{1}{(1+i)^2}=\frac{-1}{1+i}=-v.</math> '''References''' {{cite web |url=https://web2.uwind...") | Nov 26'23 at 17:23 | +664 |