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abb0t
 one year ago
Best ResponseYou've already chosen the best response.0The same as it is given.

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2Have you tried to take the first, second, and third derivative? You need to try to spot a pattern.

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2Do you know how to find the derivative?

hedmanav
 one year ago
Best ResponseYou've already chosen the best response.0no, not really sure how to find a derivative

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2So you need to know product rule, chain rule, derivative of the exponential function, and derivative of x here.

KeithAfasCalcLover
 one year ago
Best ResponseYou've already chosen the best response.0\[\eqalign{ &f(x)=xe^{x} \\ &f'(x)=f^{(1)}(x)=x\frac{d}{dx}e^{x}+e^{x}\frac{d}{dx}x=xe^{x}+e^{x}=f(x)+e^{x}\\ &f''(x)=f^{(2)}(x)=\frac{d}{dx}f^{(1)}(x)e^{x}=(xe^{x}+e^{x})+e^{x}=xe^{x}=f(x) \\ &f^{(3)}(x)=f'(x)=f^{(1)}(x) \\ &f^{(4)}(x)=f^{(2)}(x) \\ &f^{(5)}(x)=f^{(3)}(x)=f^{(1)} (x) \\ &... \\}\] 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
 one year ago
Best ResponseYou've already chosen the best response.2\[(uv)'=u'v+uv'\] \[(e^x)'=e^x\] \[(x)'=1 \text{ since the slope of x is 1.}\] ....uhh....no...I'm not getting the original function is equal to the 1600th derivative of the function

KeithAfasCalcLover
 one year ago
Best ResponseYou've already chosen the best response.0Haha I may have made a mistake. "DARN...after all that work" ;)

KeithAfasCalcLover
 one year ago
Best ResponseYou've already chosen the best response.0Lets see...

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2I 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
 one year ago
Best ResponseYou've already chosen the best response.2You didn't find the derivative of exp(x)

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2Or you did but you switched back.

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2You see what I mean?

KeithAfasCalcLover
 one year ago
Best ResponseYou've already chosen the best response.0Yes haha thank you. Ill correct it. I did over look that lol

KeithAfasCalcLover
 one year ago
Best ResponseYou've already chosen the best response.0\[\eqalign{ &f(x)=xe^{x} \\ &f'(x)=f^{(1)}(x)=x\frac{d}{dx}e^{x}+e^{x}\frac{d}{dx}x=xe^{x}+e^{x}=e^{x}(x1)\\ &\eqalign{&f''(x)=f^{(2)}(x)=\frac{d}{dx}f^{(1)}(x)+\frac{d}{dx}e^{x}=(xe^{x}+e^{x})e^{x} \\ &=xe^{x}2e^{x}=e^{x}(x2) \\ } \\ &\eqalign{&f^{(3)}(x)=[f^{(2)}(x)]'=e^{x}\frac{d}{dx}(x2)+(x2)\frac{d}{dx}e^{x} \\ &=e^{x}(x2)e^{x}=e^{x}+2e^{x}xe^{x}=3e^{x}xe^{x}=e^{x}(x3) \\} \\ &... \\} \] So we obtained the following : \[\eqalign{ &f^{(1)}(x)=e^{x}(x1) \\ &f^{(2)}(x)=e^{x}(x2) \\ &f^{(3)}(x)=e^{x}(x3) \\ }\]

KeithAfasCalcLover
 one year ago
Best ResponseYou've already chosen the best response.0So 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: \[f^{(n)}(x)=(1)^ne^{x}(xn);\phantom{..}n\in N\] 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 \(n=1600\): \[\eqalign{f^{(1600)}(x) &=(1)^{1600}e^{x}(x1600) \\ &=(1)e^{x}(x1600) \\ &=e^{x}(x1600) }\]
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