Question #53ac5

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Answer 1 · 2017-02-27T16:59:48

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:

$f(x) = xcosx$

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

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:

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