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

Jul 27, 2015

$\frac{d \left(\frac{x}{{e}^{x}} \ln \left(x\right)\right)}{\mathrm{dx}} = \frac{1 + \left(1 - x\right) \ln \left(x\right)}{{e}^{x}}$

#### Explanation:

$\frac{x}{{e}^{x}} \ln \left(x\right)$ is of the form $F \left(x\right) \cdot G \left(x\right)$

An equation of this form is deived like this:

$\frac{d \left(F \left(x\right) \cdot G \left(x\right)\right)}{\mathrm{dx}} = \frac{\mathrm{dF} \left(x\right)}{\mathrm{dx}} \cdot G \left(x\right) + F \left(x\right) \cdot \frac{\mathrm{dG} \left(x\right)}{\mathrm{dx}}$

Furthermore, $\frac{x}{{e}^{x}}$ is of the form $\frac{A \left(x\right)}{B \left(x\right)}$

An equation of this form is derived like this:

$\frac{d \frac{A \left(x\right)}{B \left(x\right)}}{\mathrm{dx}} = \frac{\frac{\mathrm{dA} \left(x\right)}{\mathrm{dx}} \cdot B \left(x\right) - A \left(x\right) \cdot \frac{\mathrm{dB} \left(x\right)}{\mathrm{dx}}}{B {\left(x\right)}^{2}}$

Knowing that:

$A \left(x\right) = x \rightarrow \frac{\mathrm{dA} \left(x\right)}{\mathrm{dx}} = 1$

$B \left(x\right) = {e}^{x} \rightarrow \frac{\mathrm{dB} \left(x\right)}{\mathrm{dx}} = {e}^{x}$

$G \left(x\right) = \ln \left(x\right) \rightarrow \frac{\mathrm{dG} \left(x\right)}{\mathrm{dx}} = \frac{1}{x}$

Therefore:

$\frac{d \left(\frac{x}{{e}^{x}} \ln \left(x\right)\right)}{\mathrm{dx}} = \frac{{e}^{x} - x {e}^{x}}{{e}^{2 x}} \ln \left(x\right) + \frac{x}{x {e}^{x}}$

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