How do you differentiate #f(x)= (1 + sin^2x)/(1 - sin2x) # using the quotient rule?

1 Answer

Answer:

#(dy)/(dx)=(2cosxsinx)/(1-sin2x)+(2cos2x(1+sin^2x))/(1-sin2x)^2#

Explanation:

We must recall two things to make this simpler. First, #f (x)=g (x)/(h (x)) -> f'(x) = (hg'-h'g)/h^2#.

Here, #g(x)=1+sin^2x=1+(sinx)^2#

#g'(x)=d/(dx)[1]+d/(dx)[(sinx)^2]#
#g'(x)=2cosxsinx#

#h(x)=1-sin(2x)#
#h'(x)=d/(dx)[1]-d/(dx)[sin2x]#
#h'(x)=-2cos2x#

#(dy)/(dx)=(2cosxsinx(1-sin2x)+2cos2x(1+sin^2x))/(1-sin2x)^2#

#(dy)/(dx)=(2cosxsinx)/(1-sin2x)+(2cos2x(1+sin^2x))/(1-sin2x)^2#