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

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

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

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

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

KeithAfasCalcLover Group TitleBest 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 \(f^{(n)}(x)=f(x)\)
 one year ago

myininaya Group TitleBest 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
 one year ago

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

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

myininaya Group TitleBest ResponseYou've already chosen the best response.2
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)
 one year ago

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

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

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

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

KeithAfasCalcLover Group TitleBest 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 : \[\eqalign{ &f^{(1)}(x)=e^{x}(x1) \\ &f^{(2)}(x)=e^{x}(x2) \\ &f^{(3)}(x)=e^{x}(x3) \\ }\]
 one year ago

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