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Question Number 95760 by  M±th+et+s last updated on 27/May/20

(d/(d(x)))(W(x))=? W(x) is lambert W function

$$\frac{{d}}{{d}\left({x}\right)}\left({W}\left({x}\right)\right)=? \\ $$$$\:{W}\left({x}\right)\:{is}\:{lambert}\:{W}\:{function} \\ $$

Answered by prakash jain last updated on 27/May/20

Do you mean W_0 (x) by W(x)? y=xe^x x=W(x)e^(W(x)) ln x=ln W(x)+W(x) (1/x)=(1/(W(x)))(d/dx)W(x)+(d/dx)W(x) (1/x)=((1+W(x))/(W(x)))∙((dW(x))/(d(x))) (d/dx)W(x)=((W(x))/(x(1+W(x))))

$$\mathrm{Do}\:\mathrm{you}\:\mathrm{mean}\:{W}_{\mathrm{0}} \left({x}\right)\:\mathrm{by}\:{W}\left({x}\right)? \\ $$$${y}={xe}^{{x}} \\ $$$${x}={W}\left({x}\right){e}^{{W}\left({x}\right)} \\ $$$$\mathrm{ln}\:{x}=\mathrm{ln}\:{W}\left({x}\right)+{W}\left({x}\right) \\ $$$$\frac{\mathrm{1}}{{x}}=\frac{\mathrm{1}}{{W}\left({x}\right)}\frac{{d}}{{dx}}{W}\left({x}\right)+\frac{{d}}{{dx}}{W}\left({x}\right) \\ $$$$\frac{\mathrm{1}}{{x}}=\frac{\mathrm{1}+{W}\left({x}\right)}{{W}\left({x}\right)}\centerdot\frac{{dW}\left({x}\right)}{{d}\left({x}\right)} \\ $$$$\frac{{d}}{{dx}}{W}\left({x}\right)=\frac{{W}\left({x}\right)}{{x}\left(\mathrm{1}+{W}\left({x}\right)\right)} \\ $$

Commented by  M±th+et+s last updated on 27/May/20

yes sir. thank you

$${yes}\:{sir}.\:{thank}\:{you} \\ $$

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