Guest Posted May 16, 2010 Report Share Posted May 16, 2010 f(x)=sin(x*sqrt(3)/2)/(e^(x/2)) f'(x)=(sqrt(3)*cos(x*sqrt(3)/2)-sin(x*sqrt(3)/2))/(2e^(x/2)) f''(x)=(sqrt(3)*-cos(x*sqrt(3)/2)-sin(x*sqrt(3)/2))/(2e^(x/2)) f'''(x)=sin(x*sqrt(3)/2)/(e^(x/2)) thus, f(x)is its own third, sixth, ninth, etc... derivative. note that: sin(x)=sin(x) cos(x)=sin(x+pi/2) -sin(x)=sin(x+2*pi/2) -cos(x)=sin(x+3*pi/2) find a formula for the nth derivative of f(x) this may help:http://www.wolframalpha.com/input/?i=e%5E%28-x%2F2%29+sin%28%28sqrt%283%29+x%29%2F2%29 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted May 17, 2010 Report Share Posted May 17, 2010 Shouldn't the Nth derivative be equal to N mod 3 since every 3rd derivative takes you back to the beginning Quote Link to comment Share on other sites More sharing options...
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Guest
f(x)=sin(x*sqrt(3)/2)/(e^(x/2))
f'(x)=(sqrt(3)*cos(x*sqrt(3)/2)-sin(x*sqrt(3)/2))/(2e^(x/2))
f''(x)=(sqrt(3)*-cos(x*sqrt(3)/2)-sin(x*sqrt(3)/2))/(2e^(x/2))
f'''(x)=sin(x*sqrt(3)/2)/(e^(x/2))
thus, f(x)is its own third, sixth, ninth, etc... derivative.
note that:
sin(x)=sin(x)
cos(x)=sin(x+pi/2)
-sin(x)=sin(x+2*pi/2)
-cos(x)=sin(x+3*pi/2)
find a formula for the nth derivative of f(x)
this may help:http://www.wolframalpha.com/input/?i=e%5E%28-x%2F2%29+sin%28%28sqrt%283%29+x%29%2F2%29
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