How do you find the derivative of 3e^ (-3/x)?

Jun 29, 2016

$= \frac{9}{x} ^ 2 {e}^{- \frac{3}{x}}$

Explanation:

$\frac{d}{\mathrm{dx}} 3 {e}^{- \frac{3}{x}}$

a few thoughts first

$\frac{d}{\mathrm{dx}} \alpha f \left(x\right) = \alpha \frac{d}{\mathrm{dx}} f \left(x\right)$

and $\frac{d}{\mathrm{dx}} {e}^{g \left(x\right)} = g ' \left(x\right) {e}^{g \left(x\right)}$ by the chain rule

so here we can say that

$\frac{d}{\mathrm{dx}} 3 {e}^{- \frac{3}{x}}$
$= 3 \frac{d}{\mathrm{dx}} {e}^{- \frac{3}{x}}$
$= 3 \frac{d}{\mathrm{dx}} \left(- \frac{3}{x}\right) {e}^{- \frac{3}{x}}$

$= 3 \frac{d}{\mathrm{dx}} \left(- 3 {x}^{- 1}\right) {e}^{- \frac{3}{x}}$

$= 3 \left(- 1\right) \left(- 3 {x}^{- 1 - 1}\right) {e}^{- \frac{3}{x}}$ by the power rule

$= 3 \cdot 3 {x}^{- 2} {e}^{- \frac{3}{x}}$

$= \frac{9}{x} ^ 2 {e}^{- \frac{3}{x}}$