Revision as of 17:51, 26 November 2023 by Admin (Created page with "Find the first derivative with respect to <math>i</math> of <math>f(i)=\frac{\delta}{d}</math>. <ul class="mw-excansopts"><li>0</li><li><math>\frac{i-\delta}{i^2}</math></li><li><math>\frac{d^2}{i^2}\left(1-\frac{i}{\delta}\right)</math></li><li><math>\frac{\delta-i}{i^2}</math></li><li><math>\frac{d^2}{i^2}\left(\frac{i}{\delta}-1\right)</math></li></ul> '''References''' {{cite web |url=https://web2.uwindsor.ca/math/hlynka/392oldtests.html |last=Hlynka |first=Myron |...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
ABy Admin
Nov 26'23

Exercise

Find the first derivative with respect to [math]i[/math] of [math]f(i)=\frac{\delta}{d}[/math].

  • 0
  • [math]\frac{i-\delta}{i^2}[/math]
  • [math]\frac{d^2}{i^2}\left(1-\frac{i}{\delta}\right)[/math]
  • [math]\frac{\delta-i}{i^2}[/math]
  • [math]\frac{d^2}{i^2}\left(\frac{i}{\delta}-1\right)[/math]

References

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

ABy Admin
Nov 26'23

Solution: B

[math]\delta=\ln (1+i)[/math] and [math]d=i /(1+i)=1-1 /(1+i)[/math]. Thus

[[math]] \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

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

00