Given cos (x) cos(2x) cos(3x) let d/dx cos x = -sin x d^2/dx^2 cos x = -cos x d^3/dx^3 cos x = sin x and then things repeat, you get the same sequence of derivatives after that. If n is odd the nth derivative is: (-1)^( (n+1)/2 ) sin x And if n is even (-1)^( n/2 ) cos x d/dx cos 2x = -2 sin 2x So the only difference is in this case we get an extra 2 each step... d^2/dx^2 cos 2x = -4 cos 2x d^3/dx^3 cos 2x = 8 sin 2x d^4/dx^4 cos 2x = 16 cos 2x The sequence is predictable, the sequence of -sin 2x, -cos 2x, sin 2x, cos 2x, coupled with the powers of two (2, 2*2, 2*2*2, 2*2*2*2, ... ). And if instead we started with cos 3x, then the sequence would have had coefficients of 3,3*3,3*3*3,etc So by this pattern, the nth derivative of cos ax would be: If n is odd, (-1)^( (n+1)/2 ) a^n sin x And if n is even (-1)^( n/2 ) a^n cos x " Understand more about Finding the Nth Term -
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