Find a general formula for f^(n)(x) if f(x) = 8xe^−x

Question

zepdrix · Answer

Hey cassie :)
Have you tried taking a few derivatives?
You might see a pattern!

zepdrix · Answer

$$\Large\rm f(x)=8xe^{-x}$$$$\Large\rm f'(x)=8e^{-x}-8x e^{-x}$$The trick is to make sure you're factoring each time.$$\Large\rm f'(x)=8e^{-x}(1-x)$$

geerky42 · Answer

Here's how I see the pattern: $$f(x)=8xe^{-x} \~\f'(x) = 8e^{-x}-8xe^{-x} = 8e^{-x}-f(x)\~\f''(x) = -8e^{-x}-f'(x) = -8e^{-x} - [8e^{-x}-f(x)] = 2(-8e^{-x})+f(x)\~\f'''(x)=8e^{-x}-f''(x) = 8e^{-x}-[2(-8e^{-x})+f(x)] = 3(8e^{-x})-f(x)$$

To make pattern more clear: $$f(x)=0(-8e^{-x}) + f(x)\~\f'(x) = 1(8e^{-x})-f(x)\~\f''(x) =2(-8e^{-x})+f(x)\~\f'''(x)= 3(8e^{-x})-f(x)\~\f''''(x) = 4(-8e^{-x})+f(x)$$See the pattern, right?

anonymous · Answer

ah yes thank you I was having difficulty finding the pattern!

geerky42 · Answer

Welcome :)