Simplify the expression

[[math]]\left(\frac{d}{d v} \delta\right)\left(\frac{d}{d i} d\right)[[/math]]

• [math]-v[/math]
• [math]-v^3[/math]
• 1
• [math]v[/math]
• [math]v^3[/math]

References

Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.

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.