## hedmanav 2 years ago what is the 1600th derivative of f(x)=xe^(-x)?

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1. abb0t

The same as it is given.

2. myininaya

Have you tried to take the first, second, and third derivative? You need to try to spot a pattern.

3. myininaya

Do you know how to find the derivative?

4. hedmanav

no, not really sure how to find a derivative

5. myininaya

So you need to know product rule, chain rule, derivative of the exponential function, and derivative of x here.

6. KeithAfasCalcLover

\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 $$f^{(n)}(x)=f(x)$$

7. 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

8. KeithAfasCalcLover

Haha I may have made a mistake. "DARN...after all that work" ;)

9. KeithAfasCalcLover

Lets see...

10. 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)

11. myininaya

You didn't find the derivative of exp(-x)

12. myininaya

Or you did but you switched back.

13. myininaya

You see what I mean?

14. KeithAfasCalcLover

Yes haha thank you. Ill correct it. I did over look that lol

15. KeithAfasCalcLover

\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) \\ }

16. KeithAfasCalcLover

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) }