#f(x) = sin^2x/cos^2x#
Let #f(x) = g(x)/(h(x))#, so that #g(x) = sin^2x# and #h(x) = cos^2x#. Then #f'(x) = (g'(x) xx h(x) - g(x) xx h'(x))/(h(x))^2#.
Let's differentiate #g(x)# and #h(x)#.
#g(x) = (sinx)(sinx)#
#g'(x) = cosxsinx + cosxsinx#
#g'(x) = 2cosxsinx#
#g'(x) = sin2x#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#h'(x) = (cosx)(cosx)#
#h'(x) = -sinxcosx - sinxcosx#
#h'(x) = -2sinxcosx#
#h'(x) = -(2sinxcosx)#
#h'(x) = -sin2x#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#f'(x) = (g'(x) xx h(x) - g(x) xx h'(x))/(h(x))^2#.
#f'(x) = ((sin2x xx cos^2x) - (-sin2x xx sin^2x))/(cos^2x)^2#
#f'(x) = ((2sinxcosx xx cos^2x) - (-2sinxcosx xx sin^2x))/(cos^2x)^2#
#f'(x) = ((2sinxcos^3x) - (-2sin^3xcosx))/(cos^4x)#
#f'(x) = (2sinxcos^3x + 2sin^3xcosx)/(cos^4x)#
#f'(x) = (2sinxcosx(cos^2x + sin^2x))/cos^4x#
#f'(x) = (2sinx)/cos^3x#
#f'(x) = 2cotxsec^2x#
Hopefully this helps!
