Find the 1000th derivative of f(x) = x(e^-x).
How do you solve this using derivative rules?

Question

myininaya · Answer

so we need to see if we can find a pattern

myininaya · Answer

$$f^{(n)}(x)=(x e^{-x})^{(n)}$$
for n=1:  
$$f^{(1)}(x)=e^{-x}+x(-e^{-x})=e^{-x}-xe^{-x}$$
for n=2:
$$f^{(2)}(x)=-e^{-x}-(e^{-x}+x(-e^{-x}))=-e^{-x}-e^{-x}+xe^{-x}=-2e^{-x}+xe^{-x}$$
for n=3:
$$f^{(3)}(x)=-2(-e^{-x})+(e^{-x}+x(-e^{-x}))=3e^{-x}-xe^{-x}$$
for n=4:
$$f^{(4)}(x)=3(-e^{-x})-(e^{-x}+x(-e^{-x}))=-4e^{-x}+xe^{-x}$$

myininaya · Answer

its alternating right?
and the coefficient of e^{-x} is increasing by 1 each time

myininaya · Answer

$$f^{(n)}(x)=(-1)^{n+1} \cdot n \cdot e^{-x}+(-1)^{n} \cdot x \cdot e^{-x}, n \ge 1, n \in \mathbb{Z}$$

myininaya · Answer

now you should have it now since i made you a formula for n-th derivative

he66666 · Answer

Alright, thanks so much!