# How do you find the derivative of  y = x^2e^(-1/x)?

Aug 4, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(1 + 2 x\right) {e}^{- \frac{1}{x}}$.

#### Explanation:

$y = {x}^{2} {e}^{- \frac{1}{x}}$

By the Product Rule, we get,

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

$= {x}^{2} \cdot {e}^{- \frac{1}{x}} \cdot \frac{d}{\mathrm{dx}} \left(- \frac{1}{x}\right) + 2 x \cdot {e}^{- \frac{1}{x}}$.......[Chain Rule]

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

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \left(1 + 2 x\right) {e}^{- \frac{1}{x}}$.