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Question Number 4915 by Yozzii last updated on 21/Mar/16
ln(((d{y(x)})/dx))=(d/dx)(ln{y(x)})  y(x)=?
$${ln}\left(\frac{{d}\left\{{y}\left({x}\right)\right\}}{{dx}}\right)=\frac{{d}}{{dx}}\left({ln}\left\{{y}\left({x}\right)\right\}\right) \\ $$$${y}\left({x}\right)=? \\ $$
Commented by Yozzii last updated on 21/Mar/16
y^′ =((d{y(x)})/dx)  ⇒lny^′ =(d/dx)(ln{y(x)})=(y^′ /(y(x)))  y(x)lny^′ =y^′   (y^′ )^(y(x)) =e^y^′    (y^′ )^((y(x))/y^′ ) =e
$${y}^{'} =\frac{{d}\left\{{y}\left({x}\right)\right\}}{{dx}} \\ $$$$\Rightarrow{lny}^{'} =\frac{{d}}{{dx}}\left({ln}\left\{{y}\left({x}\right)\right\}\right)=\frac{{y}^{'} }{{y}\left({x}\right)} \\ $$$${y}\left({x}\right){lny}^{'} ={y}^{'} \\ $$$$\left({y}^{'} \right)^{{y}\left({x}\right)} ={e}^{{y}^{'} } \\ $$$$\left({y}^{'} \right)^{\frac{{y}\left({x}\right)}{{y}^{'} }} ={e} \\ $$$$ \\ $$
Commented by 123456 last updated on 21/Mar/16
y ln y′=y′  c−x=(1/(W(−(1/y))))−ln W (−(1/y))  by:wolf :v
$${y}\:\mathrm{ln}\:{y}'={y}' \\ $$$${c}−{x}=\frac{\mathrm{1}}{{W}\left(−\frac{\mathrm{1}}{{y}}\right)}−\mathrm{ln}\:{W}\:\left(−\frac{\mathrm{1}}{{y}}\right) \\ $$$$\mathrm{by}:\mathrm{wolf}\::{v} \\ $$
Commented by Yozzii last updated on 21/Mar/16
I found that answer from wolfphramα  but I still didn′t understand its derivation.  How is the Lambert W function used?  This math area is alien to me. I just  created this question for fun to see  what such function y exists.
$${I}\:{found}\:{that}\:{answer}\:{from}\:{wolfphram}\alpha \\ $$$${but}\:{I}\:{still}\:{didn}'{t}\:{understand}\:{its}\:{derivation}. \\ $$$${How}\:{is}\:{the}\:{Lambert}\:{W}\:{function}\:{used}? \\ $$$${This}\:{math}\:{area}\:{is}\:{alien}\:{to}\:{me}.\:{I}\:{just} \\ $$$${created}\:{this}\:{question}\:{for}\:{fun}\:{to}\:{see} \\ $$$${what}\:{such}\:{function}\:{y}\:{exists}. \\ $$
Commented by prakash jain last updated on 22/Mar/16
I remember a similar question using  log product function earlier.
$$\mathrm{I}\:\mathrm{remember}\:\mathrm{a}\:\mathrm{similar}\:\mathrm{question}\:\mathrm{using} \\ $$$$\mathrm{log}\:\mathrm{product}\:\mathrm{function}\:\mathrm{earlier}. \\ $$

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