Question Number 95760 by M±th+et+s last updated on 27/May/20
$$\frac{{d}}{{d}\left({x}\right)}\left({W}\left({x}\right)\right)=? \\ $$$$\:{W}\left({x}\right)\:{is}\:{lambert}\:{W}\:{function} \\ $$
Answered by prakash jain last updated on 27/May/20
$$\mathrm{Do}\:\mathrm{you}\:\mathrm{mean}\:{W}_{\mathrm{0}} \left({x}\right)\:\mathrm{by}\:{W}\left({x}\right)? \\ $$$${y}={xe}^{{x}} \\ $$$${x}={W}\left({x}\right){e}^{{W}\left({x}\right)} \\ $$$$\mathrm{ln}\:{x}=\mathrm{ln}\:{W}\left({x}\right)+{W}\left({x}\right) \\ $$$$\frac{\mathrm{1}}{{x}}=\frac{\mathrm{1}}{{W}\left({x}\right)}\frac{{d}}{{dx}}{W}\left({x}\right)+\frac{{d}}{{dx}}{W}\left({x}\right) \\ $$$$\frac{\mathrm{1}}{{x}}=\frac{\mathrm{1}+{W}\left({x}\right)}{{W}\left({x}\right)}\centerdot\frac{{dW}\left({x}\right)}{{d}\left({x}\right)} \\ $$$$\frac{{d}}{{dx}}{W}\left({x}\right)=\frac{{W}\left({x}\right)}{{x}\left(\mathrm{1}+{W}\left({x}\right)\right)} \\ $$
Commented by M±th+et+s last updated on 27/May/20
$${yes}\:{sir}.\:{thank}\:{you} \\ $$