How do you find #f^37x# given #f(x)=cos3x#?

For problems of this kind, they want you to find a pattern for the derivatives.

By the chain rule, the first derivative is:

#f^1(x) = -3sin(3x)#

The second derivative is :

#f^2(x) = -9cos(3x)#

The third derivative is:

#f^3(x) = 27sin(3x)#

The fourth derivative is:

#f^4(x) = 81cos(3x)#

The fifth derivative is:

#f^5(x) = -243sin(3x)#

You get the pattern. If the derivative is even-numbered (example #f^4(x)#), then it ends in #cos3x#. If it is odd, (example #f^29(x)#), it ends in #sin(3x)#.

As for the sign of the derivative, it does two negative, followed by two positive, followed by two negative et.cetera. You can use an arithmetic sequence to figure out whether #f^37(x)# will have a negative sign or a positive sign. Set the sequence to #36#. We have common difference #5#, and we will know the sign of #f^37(x)# if #n# is an integer.

#36 = 1 + (n - 1)5#

#36 = 1 + 5n - 5#

#40 = 5n#

#n = 8#

Therefore, #f^37(x)# is negative.

Finally, we determine the coefficient of #f^37(x)# using a geometric sequence. If you disregard the signs, you will notice that each derivative after the first has a coefficient #3# times higher than the previous. Therefore:

#t_n = a * r^(n - 1)#

#t_37 = 3 * 3^(36)#

#t_37= 3^37#

As you can imagine, this is a massive number. Therefore, we'll have to keep it in the form that it is above.

Hence, #f^37(x) = -3^37sin(3x)#

Hopefully this helps!