Question

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

Answer 1

`(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)`