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

Question

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

William Abdalla · Accepted Answer

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

Christopher Agresta · Answer

undefined To differentiate the function $ f(x) = \frac{{1 + \sin^2(x)}}{{1 - \sin(2x)}} $ using the quotient rule, follow these steps:

1. Identify $ u $ and $ v $ as the numerator and denominator functions, respectively.
2. Apply the quotient rule: $ \frac{{d}}{{dx}}\left(\frac{{u}}{{v}}\right) = \frac{{v \cdot \frac{{du}}{{dx}} - u \cdot \frac{{dv}}{{dx}}}}{{v^2}} $.
3. Differentiate $ u $ and $ v $ separately.
4. Substitute the derivatives and the original functions into the quotient rule formula.
5. Simplify the expression if necessary.

Using these steps, you can find the derivative of $ f(x) $.