Revision as of 17:23, 26 November 2023 by 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...")
Exercise
ABy Admin
Nov 26'23
Answer
Solution A
[[math]]
\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]]\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
Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.