Asimov Pupil Posted May 6, 2005 Share Posted May 6, 2005 sorry to bother everyone but i found a question to which i never learned process so i wonder if anyone can tell me da rules or explain it to me [math]f(x)=e^{2x)[/math] find [math]f^6(0)[/math] though i think this is a fundamental technique I must have missed it when they were teaching it. so i checked the book and it said the answer was 64 now the only way i could get it to equal that, is to find the derivative [math]\frac{dy}{dx}=2e^{2x}[/math] then solve f(0) [math]f(0)=2[/math] then i thought that [math]f^6(0)=(f(0))^6[/math] so i did [math]2^6=64[/math] i'm not sure if my process is right (which i think it isn't cuz it didn't say [math](f^6(0))'[/math]) someone please explain thx Link to comment Share on other sites More sharing options...
fuhrerkeebs Posted May 6, 2005 Share Posted May 6, 2005 Your process isn't right, but your answer is. The nth derivative of e2x is 2ne2x. Using that, just plug in n=6 and you get 26=64. Link to comment Share on other sites More sharing options...
Asimov Pupil Posted May 6, 2005 Author Share Posted May 6, 2005 oh ok Link to comment Share on other sites More sharing options...
davesbird Posted May 6, 2005 Share Posted May 6, 2005 [math]f(x)=e^{2x)[/math] find [math]f^6(0)[/math] When you have [math]f^6(0)[/math] this means the 6th derivative of f. It is actually more commonly written as [math]f^{(6)}(0)[/math] to show that it is a derivative of f as opposed to a power of f. Link to comment Share on other sites More sharing options...
Asimov Pupil Posted May 6, 2005 Author Share Posted May 6, 2005 thank you Link to comment Share on other sites More sharing options...
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