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!
