Question

what is the 1600th derivative of f(x)=xe^(-x)?

Answer 1

The same as it is given.

Answer 2

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

Answer 3

Do you know how to find the derivative?

Answer 4

no, not really sure how to find a derivative

Answer 5

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

Answer 6

\[\eqalign{
&f(x)=xe^{-x} \\
&f'(x)=f^{(1)}(x)=x\frac{d}{dx}e^{-x}+e^{-x}\frac{d}{dx}x=-xe^{-x}+e^{-x}=-f(x)+e^{-x}\\
&f''(x)=f^{(2)}(x)=-\frac{d}{dx}f^{(1)}(x)-e^{-x}=-(-xe^{-x}+e^{-x})+e^{-x}=xe^{-x}=f(x) \\
&f^{(3)}(x)=f'(x)=f^{(1)}(x) \\
&f^{(4)}(x)=f^{(2)}(x) \\
&f^{(5)}(x)=f^{(3)}(x)=f^{(1)}
(x) \\
&... \\}\]
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 \(f^{(n)}(x)=f(x)\)

Answer 7

\[(uv)'=u'v+uv'\]
\[(e^x)'=e^x\]
\[(x)'=1 \text{ since the slope of x is 1.}\]


....uhh....no...I'm not getting the original function is equal to the 1600th derivative of the function

Answer 8

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

Answer 9

Lets see...

Answer 10

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)

Answer 11

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

Answer 12

Or you did but you switched back.

Answer 13

You see what I mean?

Answer 14

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

Answer 15

\[\eqalign{
&f(x)=xe^{-x} \\
&f'(x)=f^{(1)}(x)=x\frac{d}{dx}e^{-x}+e^{-x}\frac{d}{dx}x=-xe^{-x}+e^{-x}=-e^{-x}(x-1)\\
&\eqalign{&f''(x)=f^{(2)}(x)=-\frac{d}{dx}f^{(1)}(x)+\frac{d}{dx}e^{-x}=-(-xe^{-x}+e^{-x})-e^{-x} \\
&=xe^{-x}-2e^{-x}=e^{-x}(x-2) \\ } \\
&\eqalign{&f^{(3)}(x)=[f^{(2)}(x)]'=e^{-x}\frac{d}{dx}(x-2)+(x-2)\frac{d}{dx}e^{-x} \\
&=e^{-x}-(x-2)e^{-x}=e^{-x}+2e^{-x}-xe^{-x}=3e^{-x}-xe^{-x}=-e^{-x}(x-3) \\} \\
&... \\}
\]
So we obtained the following :
\[\eqalign{
&f^{(1)}(x)=-e^{-x}(x-1) \\
&f^{(2)}(x)=e^{-x}(x-2) \\
&f^{(3)}(x)=-e^{-x}(x-3) \\
}\]

Answer 16

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 \(n=1600\):
\[\eqalign{f^{(1600)}(x)
&=(-1)^{1600}e^{-x}(x-1600) \\
&=(1)e^{-x}(x-1600) \\
&=e^{-x}(x-1600)
}\]