## hedmanav Group Title what is the 1600th derivative of f(x)=xe^(-x)? 11 months ago 11 months ago

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1. abb0t Group Title

The same as it is given.

2. myininaya Group Title

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

3. myininaya Group Title

Do you know how to find the derivative?

4. hedmanav Group Title

no, not really sure how to find a derivative

5. myininaya Group Title

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

6. KeithAfasCalcLover Group Title

\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 Group Title

$(uv)'=u'v+uv'$ $(e^x)'=e^x$ $(x)'=1 \text{ since the slope of x is 1.}$ ....uhh....no...I'm not getting the original function is equal to the 1600th derivative of the function

8. KeithAfasCalcLover Group Title

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

9. KeithAfasCalcLover Group Title

Lets see...

10. myininaya Group Title

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 Group Title

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

12. myininaya Group Title

Or you did but you switched back.

13. myininaya Group Title

You see what I mean?

14. KeithAfasCalcLover Group Title

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

15. KeithAfasCalcLover Group Title

\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 Group Title

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) }