Help with calculus Find the n th derivative of y, if y = (1-x^2)cosx So we're asked for an explicit formula for the n th drivative. Thanks
These kind of problem always has a pattern. I would keep differentiating it until there is a pattern.
I can get the pattern for cosx, the problem is that x^2cosx, the derivative just gets larger end larger
\[f(x)=(1-x^2)cosx\] \[f'(x)=(cosx-x^2cosx)'=-sinx-2xcosx+x^2sinx=-2xcosx-(1-x^2)sinx\] \[f''(x)=(-2xcosx-sinx+x^2sinx)'=-2cosx+2xsinx-cosx+2xsinx+\] \[x^2cosx\]
The second derivative simplifies farther to \[4xsinx-(3-x^2)cosx\]
The third derivative skipping the calculation, is \[6xcosx+(7-x^2)sinx\]
The fourth derivative is \[-8xsinx+(13-x^2)cosx\]
The fifth derivative is \[-10xcosx-(21-x^2)sinx\]
See a pattern?
Maybe f^n(x) = 2nxsin(x+n*pi/2)+(somenumber-x^2)cos(x+n*pi/2)
but what is that number
that "somenumber" are 1,3,7,13,21...etc 1+2=3 3+4=7 7+6=13 13+8=21 ...
ok, i got it, thanks
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