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)