Question

differentiate it.
x logx/e^x
​

Answer 1

Step-by-step explanation:

We have,

$y = \frac{x log(x) }{ {e}^{x} }$

Differentiating both sides, we have,

$\frac{dy}{dx} = \frac{ {e}^{x}. \frac{d}{dx} (x log(x)) - (x log(x) ) \frac{d}{dx} ( {e}^{x} )}{ { ({e}^{x}) }^{2} }$

$\implies \frac{dy}{dx} = \frac{ {e}^{x} ( log(x) + 1) - {e}^{x} .x log(x) }{ {e}^{2x} }$

$\implies \frac{dy}{dx} = \frac{ {e}^{x}(x log(x) + log(x) + 1)}{ {e}^{2x} }$

$\implies \frac{dy}{dx} = {e}^{ - x} ((x + 1) log(x) + 1)$