anonymous
  • anonymous
what is the 1600th derivative of f(x)=xe^(-x)?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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abb0t
  • abb0t
The same as it is given.
myininaya
  • myininaya
Have you tried to take the first, second, and third derivative? You need to try to spot a pattern.
myininaya
  • myininaya
Do you know how to find the derivative?

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anonymous
  • anonymous
no, not really sure how to find a derivative
myininaya
  • myininaya
So you need to know product rule, chain rule, derivative of the exponential function, and derivative of x here.
anonymous
  • anonymous
\[\eqalign{ &f(x)=xe^{-x} \\ &f'(x)=f^{(1)}(x)=x\frac{d}{dx}e^{-x}+e^{-x}\frac{d}{dx}x=-xe^{-x}+e^{-x}=-f(x)+e^{-x}\\ &f''(x)=f^{(2)}(x)=-\frac{d}{dx}f^{(1)}(x)-e^{-x}=-(-xe^{-x}+e^{-x})+e^{-x}=xe^{-x}=f(x) \\ &f^{(3)}(x)=f'(x)=f^{(1)}(x) \\ &f^{(4)}(x)=f^{(2)}(x) \\ &f^{(5)}(x)=f^{(3)}(x)=f^{(1)} (x) \\ &... \\}\] So therefore, we can summarize the following by so: \[f^{(n)}(x)=\left\{\eqalign{ &f^{(1)}(x)=e^{-x}-f(x);\phantom{..}(n=2k+1)\wedge (k\in N) \\ &f(x)=xe^{-x};\phantom{..}(n=2k)\wedge (k\in N) \\ }\right\}\] In other words, if \(n\) in \(f^{(n)}(x)\) is odd, then \(f^{(n)}(x)=e^{-x}-f(x)\). If \(n\) in \(f^{(n)}(x)\) is even, then \(f^{(n)}(x)=f(x)\)
myininaya
  • myininaya
\[(uv)'=u'v+uv'\] \[(e^x)'=e^x\] \[(x)'=1 \text{ since the slope of x is 1.}\] ....uhh....no...I'm not getting the original function is equal to the 1600th derivative of the function
anonymous
  • anonymous
Haha I may have made a mistake. "DARN...after all that work" ;)
anonymous
  • anonymous
Lets see...
myininaya
  • myininaya
I think your line 3 you found the first derivative wrong of that one function you spotted and renamed f(x) since it did match our f(x)
myininaya
  • myininaya
You didn't find the derivative of exp(-x)
myininaya
  • myininaya
Or you did but you switched back.
myininaya
  • myininaya
You see what I mean?
anonymous
  • anonymous
Yes haha thank you. Ill correct it. I did over look that lol
anonymous
  • anonymous
\[\eqalign{ &f(x)=xe^{-x} \\ &f'(x)=f^{(1)}(x)=x\frac{d}{dx}e^{-x}+e^{-x}\frac{d}{dx}x=-xe^{-x}+e^{-x}=-e^{-x}(x-1)\\ &\eqalign{&f''(x)=f^{(2)}(x)=-\frac{d}{dx}f^{(1)}(x)+\frac{d}{dx}e^{-x}=-(-xe^{-x}+e^{-x})-e^{-x} \\ &=xe^{-x}-2e^{-x}=e^{-x}(x-2) \\ } \\ &\eqalign{&f^{(3)}(x)=[f^{(2)}(x)]'=e^{-x}\frac{d}{dx}(x-2)+(x-2)\frac{d}{dx}e^{-x} \\ &=e^{-x}-(x-2)e^{-x}=e^{-x}+2e^{-x}-xe^{-x}=3e^{-x}-xe^{-x}=-e^{-x}(x-3) \\} \\ &... \\} \] So we obtained the following : \[\eqalign{ &f^{(1)}(x)=-e^{-x}(x-1) \\ &f^{(2)}(x)=e^{-x}(x-2) \\ &f^{(3)}(x)=-e^{-x}(x-3) \\ }\]
anonymous
  • anonymous
So we can observe that there is a direct relation from the derivative number (the number in parentheses above the \(f\)) and the actual derivative. We can observe that: \[f^{(n)}(x)=(-1)^ne^{-x}(x-n);\phantom{..}n\in N\] So therefore, putting 1,2,3 in for n, you'll see that the equation holds true! So we can evaluate this new equation at \(n=1600\): \[\eqalign{f^{(1600)}(x) &=(-1)^{1600}e^{-x}(x-1600) \\ &=(1)e^{-x}(x-1600) \\ &=e^{-x}(x-1600) }\]

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