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Ostrovityanka [42]
3 years ago
5

Let f(x) = -5e^(-x/3) Find f^(6) (1)

Mathematics
1 answer:
Luba_88 [7]3 years ago
6 0

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}}.

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