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anonymous
 5 years ago
f(x)=x sin x(x^2) Find f^(2011) (0).
anonymous
 5 years ago
f(x)=x sin x(x^2) Find f^(2011) (0).

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.05.58611 degrees; but is 2011 in radians or degrees for that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I believe you have to work this using Taylor series

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0oh.... then you prolly want someone else to help you ; like the cow :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0no..... rude would be me doing something absurd and uncalled for; this is just me not being good with taylor series and suggesting you find someone smarter than me...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0like...the cow here :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh ok ok i mis understood u then . thanks for attempting

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0sorry i couldnt help out :) good luck with it tho ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Just for clarification, is it \[x \sin (x*x ^{2})\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so if you taylor expand this, you will get\[x(x^2  (x^2)^3/3!  (x^2)^5/5! + ...)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and if you plug in 0 for this you get 0 ..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well, I just used the taylor expansion of sin x and then I plugged in x^2 for that x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then multiplied the series by x right??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and because you can always factor out an x term, when you plug in x, you'll get 0, or at least this is my idea of how it works

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0"no its (x)(sin(x^2)" f(x)= x*sin(x^2) f(0) = 0. At the origin both x x^2 are zero. Thus the product of x and Sin[x^2] is zero. Refer to the attached plot, Nora_1.pdf .

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Robtobey, I think the problem isn't taking the function to the 2011 power, but rather taking the 2011 derivative
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