First of all, we can simplify the function as follows:

#f(x) = cancel(x)/(cancel(x^2)^x(1 - sinx + cosx))#

#f(x) = 1/(x - xsinx + xcosx)#

**Method 1: Through the quotient rule**

Let #f(x) = g(x)/(h(x))#. Then #g(x) = 1# and #h(x) = x - xsinx + xcosx#

#g'(x) = 0# and #h'(x) = 1 - (sinx + xcosx) + (cosx - xsinx) = 1 - sinx - xcosx + cosx - xsinx#

By the quotient rule:

#color(red)(f'(x) = (g'(x) xx h(x) - h'(x) xx g(x))/(h(x))^2#

#f'(x) = ((0)(x - xsinx + xcosx) - (1(1 - sinx - xcosx + cosx - xsinx)))/(x - xsinx + xcosx)^2#

#f'(x) = (-(1 - sinx- xcosx + cosx - xsinx))/(x - xsinx + xcosx)^2#

#f'(x) = (sinx - 1 + xcosx - cosx + xsinx)/(x - xsinx + xcosx)^2#

**Method 2: By the chain rule**

We can write #f(x) = 1/(x - xsinx + xcosx)# as #f(x) = (x - xsinx + xcosx)^-1#.

We let #y = u^-1# and #u = x - xsinx + xcosx#.

#dy/(du) = -1/(u^2)# and #(du)/dx = 1 - sinx - xcosx + cosx - xsinx# (see above).

The chain rule states:

#color(red)(dy/dx = dy/(du) xx (du)/dx#

#dy/dx = -1/u^2 xx (1 - sinx - xcosx + cosx - xsinx)#

#dy/dx = (sinx + xcosx - 1 + xsinx - cosx)/(x - xsinx + xcosx)^2#

Same answer as before!

Hopefully this helps!