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

1 Answer
Apr 26, 2018

Answer:

# 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) #