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

Question

Isaiah Allen · Accepted Answer

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

#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)#

Ezra Adams · Answer

undefined To find the derivative of $ y = \frac{x}{e^x} \ln(x) $:

Use the product rule:

$$ (uv)' = u'v + uv' $$

Where:
$$ u = \frac{x}{e^x} $$
$$ v = \ln(x) $$

$$ u' = \frac{d}{dx} \left( \frac{x}{e^x} \right) $$
$$ v' = \frac{d}{dx} \ln(x) $$

Using quotient rule for $ u' $:

$$ \frac{d}{dx} \left( \frac{x}{e^x} \right) = \frac{1 \cdot e^x - x \cdot e^x}{(e^x)^2} $$
$$ \frac{d}{dx} \left( \frac{x}{e^x} \right) = \frac{e^x - x e^x}{e^{2x}} $$

$$ \frac{d}{dx} \ln(x) = \frac{1}{x} $$

Now, apply the product rule:

$$ y' = \left( \frac{e^x - x e^x}{e^{2x}} \right) \ln(x) + \frac{x}{e^x} \cdot \frac{1}{x} $$

Simplify:

$$ y' = \frac{e^x - x e^x}{e^{2x}} \ln(x) + \frac{1}{e^x} $$

That's the derivative of $ y = \frac{x}{e^x} \ln(x) $.