Question

Question #53ac5

Answer 1

The `n`-th derivative can be evaluated recursively as:

`d^n/(dx^n) f(x) = 2 sin(x+(npi)/2) - d^(n-2)/(dx^(n-2))f(x)`

with:

`f(x) = xcosx`

`f'(x) = sin(x+pi/2) -xsinx`

Explanation:

We can start from the first derivative:

`f'(x) = d/dx (xcosx) = d/dx(x) cosx + x d/dx(cosx) = cosx -xsinx`

Differentiating again:

`(1) f''(x) = -sinx -[d/dx(x) sinx +x d/dx(sinx)] = -2sinx -f(x)`

Now we know that:

`d^n/(dx^n) sinx = sin(x+(npi)/2)`

and in particular:

`d^2/(dx^2) sinx = -sinx`

so that we can write `(1)` as:

`d^2/(dx^2) f(x) = 2 d^2/(dx^2) sinx - f(x)`

and differentiating this equation we obtain a recursive formula for the derivatives of higher order:

`d^n/(dx^n) f(x) = 2 d^n/(dx^n) sinx - d^(n-2)/(dx^(n-2))f(x)`

or:

`d^n/(dx^n) f(x) = 2 sin(x+(npi)/2) - d^(n-2)/(dx^(n-2))f(x)`