Question

What will be the solution of the mentioned problem????

If`y=cos3x`
then, `y_n=?`

Answer 1

# y_n = (d^n)/(dx^n) cos3x = { ((-1)^(n/2 ) \ 3^n \ sin 3x,n " even"), ((-1)^((n+1)/(2)) \ 3^n \ cos 3x,n " odd") :} #

Explanation:

We have:

`y = cos3x`

Using the notation `y_n` to denote the `n^(th)` derivative of `y` wrt `x`.

Differentiating once wrt `x` (using the chain rule), we get the first derivative:

`y_1 = (-sin3x)(3) = -3sin3x`

Differentiating further times we get:

`y_2 = (-3)(cos3x)(3) \ \ \ \ \ \ = -3^2cos3x`
`y_3 = (-3^2)(-sin3x)(3) = +3^3sin3x`
`y_4 = (3^3)(cos3x)(3) \ \ \ \ \ \ \ \ \ \= +3^4cos3x`
`y_5 = (3^4)(-sin3x)(3) \ \ \ \ \ \= -3^5sin3x`
`vdots`

And a clear pattern is now forming, and the `n^(th)` derivative is:

# y_n = (d^n)/(dx^n) cos3x = { ((-1)^(n/2 ) \ 3^n \ sin 3x,n " even"), ((-1)^((n+1)/(2)) \ 3^n \ cos 3x,n " odd") :} #