# Math virtual assistant

Calculators Topics Go Premium About Snapxam

# Step-by-step Solution

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## Answer

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

## Step-by-step explanation

Problem to solve:

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

Apply the product rule for differentiation: $(f\cdot g)'=f'\cdot g+f\cdot g'$, where $f=e^{-x}$ and $g=\ln\left(x\right)$

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

The derivative of the natural logarithm of a function is equal to the derivative of the function divided by that function. If $f(x)=ln\:a$ (where $a$ is a function of $x$), then $\displaystyle f'(x)=\frac{a'}{a}$

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

## Answer

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

Product rule

~ 0.5 seconds