how would i go about finding the millionth derivative for f(x)=xe^-x.
H henry200 New member Joined Sep 9, 2007 Messages 10 Oct 31, 2007 #1 how would i go about finding the millionth derivative for f(x)=xe^-x.
D Deleted member 4993 Guest Oct 31, 2007 #2 Re: millionth derivative henry200 said: how would i go about finding the millionth derivative for f(x)=xe^-x. Click to expand... f'(x) = e^(-x) - x*e^(-x) f"(x) = -e^(-x) -[e^(-x) - x*e^(-x)] = - 2*e^(-x) + x*e^(-x) f"'(x) = 2*e^(-x) + [e^(-x) - x*e^(-x)] = 3e^(-x) - x*e^(-x) <---- corrected f""(x) = -3e^(-x) - [e^(-x) - x*e^(-x)] ....see a pattern!!
Re: millionth derivative henry200 said: how would i go about finding the millionth derivative for f(x)=xe^-x. Click to expand... f'(x) = e^(-x) - x*e^(-x) f"(x) = -e^(-x) -[e^(-x) - x*e^(-x)] = - 2*e^(-x) + x*e^(-x) f"'(x) = 2*e^(-x) + [e^(-x) - x*e^(-x)] = 3e^(-x) - x*e^(-x) <---- corrected f""(x) = -3e^(-x) - [e^(-x) - x*e^(-x)] ....see a pattern!!
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Nov 1, 2007 #3 Hello, henry200! Find the millionth derivative for: \(\displaystyle \:f(x)\:=\:xe^{-x}\) Click to expand... Look for a pattern . . . \(\displaystyle \begin{array}{ccccc}f'(x) & \;= \;& x(-e^{-x})\,+\,1\cdot e^{-x} & \;= \;& -e^{-x}(x\,-\,1) \\ \\ f''(x) &\;=\;& e^{-x}(-1)\,-\,e^{-x}(1\,-\,x) & \;=\; & e^{-x}(x\,-\,2) \\ \\ f'''(x)&\;=\;&e^{-x}\cdot1\,-\,e^{-x}(x\,-\,2) &\;=\;& -e^{-x}(x\,-\,3) \\ \\ f^{(4)} &\;=\;& e^{-x}(-1)\,-\,e^{-x}(3\,-\,x) &\;=\;& e^{-x}(x\,-\,4) \\ \\ \vdots & & \vdots & & \vdots\end{array}\) . . . . . . . See it? The n<sup>th</sup> derivative is: \(\displaystyle \:f^{(n)} \;=\;(-1)^ne^{-x}(x\,-\,n)\)
Hello, henry200! Find the millionth derivative for: \(\displaystyle \:f(x)\:=\:xe^{-x}\) Click to expand... Look for a pattern . . . \(\displaystyle \begin{array}{ccccc}f'(x) & \;= \;& x(-e^{-x})\,+\,1\cdot e^{-x} & \;= \;& -e^{-x}(x\,-\,1) \\ \\ f''(x) &\;=\;& e^{-x}(-1)\,-\,e^{-x}(1\,-\,x) & \;=\; & e^{-x}(x\,-\,2) \\ \\ f'''(x)&\;=\;&e^{-x}\cdot1\,-\,e^{-x}(x\,-\,2) &\;=\;& -e^{-x}(x\,-\,3) \\ \\ f^{(4)} &\;=\;& e^{-x}(-1)\,-\,e^{-x}(3\,-\,x) &\;=\;& e^{-x}(x\,-\,4) \\ \\ \vdots & & \vdots & & \vdots\end{array}\) . . . . . . . See it? The n<sup>th</sup> derivative is: \(\displaystyle \:f^{(n)} \;=\;(-1)^ne^{-x}(x\,-\,n)\)
