Question

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

Answer 1

`f^37(x) = -3^37sin(3x)`

Explanation:

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!