Question

Find a formula for nth derivative of `f(x)=x sin x`?

Answer 1

`y^((n)) = n sin(x + ((n-1)pi)/2) + x sin(x + n(pi)/2)`

Explanation:

We seek the `n^(th)` derivative of:

`f(x) = xsinx`

Starting with the given function:

`f^((0))(x) = xsinx`

Using the product rule we compute the first derivative:

`f^((1))(x) = x(d/dxsinx)+(d/dxx)sinx`
`\ \ \ \ \ \ \ \ \ \ \ \ = xcosx+sinx`

Similarity, for the second derivative

`f^((2))(x) = x(d/dxcosx)+(d/dxx)sinx + d/dxsinx`
`\ \ \ \ \ \ \ \ \ \ \ \ = -xsinx+cosx + cosx`
`\ \ \ \ \ \ \ \ \ \ \ \ = 2cosx-xsinx`

And further derivatives:

`f^((3))(x) = -2sinx-(xcosx+sinx)`
`\ \ \ \ \ \ \ \ \ \ \ \ = -3sinx-xcosx`

`f^((4))(x) = -3cosx-(-xsinx+cosx)`
`\ \ \ \ \ \ \ \ \ \ \ \ = -4cosx+xsinx`

So, By exploiting the phase shift properties, we have:

`y^((n)) = n sin(x + ((n-1)pi)/2) + x sin(x + n(pi)/2)`