Question-148276

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

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

y(n)=e^(((n^2 +n)/(n^3 −4n^2 ))log2) ⇒(dy/dn)=(d/dn)(log2×((n^2 +n)/(n^3 −4n^2 )))y(n) =log2(((n+1)/(n^2 −4n)))^((1)) y(n) =log2(((n^2 −4n−(n+1)(2n−4))/((n^2 −4n)^2 )))y(n) =log2×((n^2 −4n−(2n^2 −2n−4))/((n^2 −4n)^2 ))y(n) =log2×((−n^2 −2n+4)/((n^2 −4n)^2 ))y(n) ⇒ (dy/dn)=−log(2)×((n^2 +2n−4)/((n^2 −4n)^2 ))y(n)

$$\mathrm{y}\left(\mathrm{n}\right)=\mathrm{e}^{\frac{\mathrm{n}^{\mathrm{2}} +\mathrm{n}}{\mathrm{n}^{\mathrm{3}} −\mathrm{4n}^{\mathrm{2}} }\mathrm{log2}} \:\Rightarrow\frac{\mathrm{dy}}{\mathrm{dn}}=\frac{\mathrm{d}}{\mathrm{dn}}\left(\mathrm{log2}×\frac{\mathrm{n}^{\mathrm{2}} \:+\mathrm{n}}{\mathrm{n}^{\mathrm{3}} −\mathrm{4n}^{\mathrm{2}} }\right)\mathrm{y}\left(\mathrm{n}\right) \\ $$$$=\mathrm{log2}\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{n}^{\mathrm{2}} −\mathrm{4n}}\right)^{\left(\mathrm{1}\right)} \mathrm{y}\left(\mathrm{n}\right) \\ $$$$=\mathrm{log2}\left(\frac{\mathrm{n}^{\mathrm{2}} −\mathrm{4n}−\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{2n}−\mathrm{4}\right)}{\left(\mathrm{n}^{\mathrm{2}} −\mathrm{4n}\right)^{\mathrm{2}} }\right)\mathrm{y}\left(\mathrm{n}\right) \\ $$$$=\mathrm{log2}×\frac{\mathrm{n}^{\mathrm{2}} −\mathrm{4n}−\left(\mathrm{2n}^{\mathrm{2}} −\mathrm{2n}−\mathrm{4}\right)}{\left(\mathrm{n}^{\mathrm{2}} −\mathrm{4n}\right)^{\mathrm{2}} }\mathrm{y}\left(\mathrm{n}\right) \\ $$$$=\mathrm{log2}×\frac{−\mathrm{n}^{\mathrm{2}} −\mathrm{2n}+\mathrm{4}}{\left(\mathrm{n}^{\mathrm{2}} −\mathrm{4n}\right)^{\mathrm{2}} }\mathrm{y}\left(\mathrm{n}\right)\:\Rightarrow \\ $$$$\frac{\mathrm{dy}}{\mathrm{dn}}=−\mathrm{log}\left(\mathrm{2}\right)×\frac{\mathrm{n}^{\mathrm{2}} +\mathrm{2n}−\mathrm{4}}{\left(\mathrm{n}^{\mathrm{2}} −\mathrm{4n}\right)^{\mathrm{2}} }\mathrm{y}\left(\mathrm{n}\right) \\ $$

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