# How do you derive y = (sinx)/[1 - sin^2(x)) using the quotient rule?

Nov 30, 2015

I found: $y ' = \frac{\cos \left(x\right) \left[1 + {\sin}^{2} \left(x\right)\right]}{{\left[1 - {\sin}^{2} \left(x\right)\right]}^{2}} =$

#### Explanation:

Given a function as $f \left(x\right) = \frac{h \left(x\right)}{g} \left(x\right)$ the Quotient Rule tells us that:

$y ' = f ' \left(x\right) = \frac{h ' \left(x\right) g \left(x\right) - h \left(x\right) g ' \left(x\right)}{{\left[g \left(x\right)\right]}^{2}}$

In your case you get:

y'=f´(x)=(cos(x)(1-sin^2(x))-sin(x)(-2sin(x)cos(x)))/([1-sin^2(x)]^2)=
$= \frac{\cos \left(x\right) - \cos \left(x\right) {\sin}^{2} \left(x\right) + 2 {\sin}^{2} \left(x\right) \cos \left(x\right)}{{\left[1 - {\sin}^{2} \left(x\right)\right]}^{2}} =$
$= \frac{\cos \left(x\right) + \cos \left(x\right) {\sin}^{2} \left(x\right)}{{\left[1 - {\sin}^{2} \left(x\right)\right]}^{2}} =$
$= \frac{\cos \left(x\right) \left[1 + {\sin}^{2} \left(x\right)\right]}{{\left[1 - {\sin}^{2} \left(x\right)\right]}^{2}} =$