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

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

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

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

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

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

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

KeithAfasCalcLover
 2 years 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
 2 years 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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