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)f1(x)=−3sin(3x)
The second derivative is :
f^2(x) = -9cos(3x)f2(x)=−9cos(3x)
The third derivative is:
f^3(x) = 27sin(3x)f3(x)=27sin(3x)
The fourth derivative is:
f^4(x) = 81cos(3x)f4(x)=81cos(3x)
The fifth derivative is:
f^5(x) = -243sin(3x)f5(x)=−243sin(3x)
You get the pattern. If the derivative is even-numbered (example f^4(x)f4(x)), then it ends in cos3xcos3x. If it is odd, (example f^29(x)f29(x)), it ends in sin(3x)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)f37(x) will have a negative sign or a positive sign. Set the sequence to 3636. We have common difference 55, and we will know the sign of f^37(x)f37(x) if nn is an integer.
36 = 1 + (n - 1)536=1+(n−1)5
36 = 1 + 5n - 536=1+5n−5
40 = 5n40=5n
n = 8n=8
Therefore, f^37(x)f37(x) is negative.
Finally, we determine the coefficient of f^37(x)f37(x) using a geometric sequence. If you disregard the signs, you will notice that each derivative after the first has a coefficient 33 times higher than the previous. Therefore:
t_n = a * r^(n - 1)tn=a⋅rn−1
t_37 = 3 * 3^(36)t37=3⋅336
t_37= 3^37t37=337
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)f37(x)=−337sin(3x)
Hopefully this helps!
