Question

Let f(x) = -5e^(-x/3)

Find f^(6) (1)

Answer 1

Let's take the first derivative:

$f'(x) = (-5e^{-x/3})' = -5 \times (e^{-x/3})' = -5e^{-x/3} \times \left(-\dfrac{1}{3}\right) = \dfrac{5}{3}e^{-x/3}.$

Notice that we can write this as:

$f'(x) = -\dfrac{f(x)}{3}.$

By taking the derivative of both sides $n$ times, we get:

$f^{(n+1)}(x) = -\dfrac{f^{(n)}}{3}.$

This means that each time you take a derivative, a factor of $-\dfrac{1}{3}$ will appear. So we conclude that:

$f^{(n)}(x) = \left(-\dfrac{1}{3}\right)^nf(x).$

Taking $n=6$ and $x=1$, we get:

$f^{(6)}(1) = \left(-\dfrac{1}{3}\right)^6 f(1) = \dfrac{-5e^{-1/3}}{729} = -\dfrac{5}{729}e^{-1/3}.$

So we finally get:

$\boxed{-\dfrac{5}{729}e^{-1/3}}.$