Question

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

Answer 1

`= 9/x^2 e^ (-3/x)`

Explanation:

`d/dx 3e^ (-3/x)`

a few thoughts first

`d/dx alpha f(x) = alpha d/dx f(x)`

and `d/dx e^{g(x)} = g'(x) e^{g(x))` by the chain rule

so here we can say that

`d/dx 3e^ (-3/x)`
`= 3 d/dx e^ (-3/x)`
`= 3 d/dx (- 3/x) e^ (-3/x)`

`= 3 d/dx (- 3x^{-1}) e^ (-3/x)`

`= 3 (-1) (- 3x^{-1-1}) e^ (-3/x)` by the power rule

`= 3 * 3x^{-2} e^ (-3/x)`

`= 9/x^2 e^ (-3/x)`