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

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Answer 1 · 2017-09-28T02:29:44

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

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

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

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

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

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

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

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