What is the derivative of #y=(x/e^x)LnX#?

1 Answer
Jul 27, 2015

#(d(x/(e^x)ln(x)))/dx=(1+(1-x)ln(x))/(e^x)#

Explanation:

#x/(e^x)ln(x)# is of the form #F(x)*G(x)#

An equation of this form is deived like this:

#(d(F(x)*G(x)))/dx=(dF(x))/dx*G(x)+F(x)*(dG(x))/dx#

Furthermore, #x/(e^x)# is of the form #(A(x))/(B(x))#

An equation of this form is derived like this:

#(d(A(x))/(B(x)))/dx=((dA(x))/dx*B(x)-A(x)*(dB(x))/dx)/(B(x)^2)#

Knowing that:

#A(x)=x rarr (dA(x))/dx=1#

#B(x)=e^x rarr (dB(x))/dx=e^x#

#G(x)=ln(x) rarr (dG(x))/dx=1/x#

Therefore:

#(d(x/(e^x)ln(x)))/dx=(e^x-xe^x)/(e^(2x))ln(x)+x/(xe^x)#

#=(1-x)/(e^x)ln(x)+1/(e^x)=(1+(1-x)ln(x))/(e^x)#