Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 148276 by 0731619 last updated on 26/Jul/21

Answered by mathmax by abdo last updated on 26/Jul/21

$$\mathrm{y}\left(\mathrm{n}\right)={\mathrm{e}}^{\frac{{\mathrm{n}}^2+\mathrm{n}}{{\mathrm{n}}^3-{4\mathrm{n}}^2}\log 2}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dn}}=\frac{\mathrm{d}}{\mathrm{dn}}\left(\log 2\times \frac{{\mathrm{n}}^2+\mathrm{n}}{{\mathrm{n}}^3-{4\mathrm{n}}^2}\right)\mathrm{y}\left(\mathrm{n}\right)$$$$=\log 2{\left(\frac{\mathrm{n}+1}{{\mathrm{n}}^2-4\mathrm{n}}\right)}^{\left(1\right)}\mathrm{y}\left(\mathrm{n}\right)$$$$=\log 2\left(\frac{{\mathrm{n}}^2-4\mathrm{n}-(\mathrm{n}+1)(2\mathrm{n}-4)}{{({\mathrm{n}}^2-4\mathrm{n})}^2}\right)\mathrm{y}\left(\mathrm{n}\right)$$$$=\log 2\times \frac{{\mathrm{n}}^2-4\mathrm{n}-({2\mathrm{n}}^2-2\mathrm{n}-4)}{{({\mathrm{n}}^2-4\mathrm{n})}^2}\mathrm{y}\left(\mathrm{n}\right)$$$$=\log 2\times \frac{-{\mathrm{n}}^2-2\mathrm{n}+4}{{({\mathrm{n}}^2-4\mathrm{n})}^2}\mathrm{y}\left(\mathrm{n}\right)\Rightarrow$$$$\frac{\mathrm{dy}}{\mathrm{dn}}=-\log \left(2\right)\times \frac{{\mathrm{n}}^2+2\mathrm{n}-4}{{({\mathrm{n}}^2-4\mathrm{n})}^2}\mathrm{y}\left(\mathrm{n}\right)$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com