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Question-151766




Question Number 151766 by mathdanisur last updated on 22/Aug/21
Commented by tabata last updated on 23/Aug/21
ln(y) ln(x)=ln(3)⇒ln(y)=((ln(3))/(ln(x)))    ⇒(y^′ /y)=−((ln(3))/(x (ln(x))^2 ))     (dy/dx) ∣y=e ⇒(dy/dx) = − ((e ln (3))/(x (ln(x))^2 )) = − ((ln(3)^e )/(x (ln(x))^2 ))    ⟨M.T ⟩
$${ln}\left({y}\right)\:{ln}\left({x}\right)={ln}\left(\mathrm{3}\right)\Rightarrow{ln}\left({y}\right)=\frac{{ln}\left(\mathrm{3}\right)}{{ln}\left({x}\right)} \\ $$$$ \\ $$$$\Rightarrow\frac{{y}^{'} }{{y}}=−\frac{{ln}\left(\mathrm{3}\right)}{{x}\:\left({ln}\left({x}\right)\right)^{\mathrm{2}} } \\ $$$$ \\ $$$$\:\frac{{dy}}{{dx}}\:\mid{y}={e}\:\Rightarrow\frac{{dy}}{{dx}}\:=\:−\:\frac{{e}\:{ln}\:\left(\mathrm{3}\right)}{{x}\:\left({ln}\left({x}\right)\right)^{\mathrm{2}} }\:=\:−\:\frac{{ln}\left(\mathrm{3}\right)^{{e}} }{{x}\:\left({ln}\left({x}\right)\right)^{\mathrm{2}} } \\ $$$$ \\ $$$$\langle{M}.{T}\:\rangle \\ $$

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