How to find the 48th derivative of the function cos2x?NB: Is there any other possible way without having to do each derivative from 1-48?

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hala718 | High School Teacher | (Level 1) Educator Emeritus

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Let us find the series of differentiating cos2x.

==> Let y= cos2x

=> y' = -2sin2x

==> y'' = -2*2cos2x = -4cos2x

==> y''' = -4*-2*sin2x = 8sin2x

==> y'''' = 8*2*cos2x = 16cos2x

Now we will find the relation of the sequence.

==> -2sin2x, -4cos2x, 8sin2x, 16cosx , -32sin2x, -64cos2x,....

an : -2sin2x, -32sin2x, .....  n= 1, 5, 9, ...

an: -4cos2x , -64cos2x, ...  n= 2, 6, 10, ...

an: 8sin2x , 128sin2x, .... n= 3, 7, 11, ...

an : 16cos2x, 256cos2x ,.. n= 4, 8, 12, ...

Now we notice that the 4th derivative belongs to the 4th series which is an: 16cos2x , 256co2x n= 4, 8, 12,16, 20, ....

==> a48 = (2)^48 *cos2x 

==>Then the 48th derivative is : 2^48 * cos2x

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