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

$f \left(x\right) = \frac{1}{\sqrt{{\sin}^{2} \left(2 - {x}^{2}\right)}} = {\left({\sin}^{2} \left(2 - {x}^{2}\right)\right)}^{- \frac{1}{2}} = {\left({\left(\sin \left(2 - {x}^{2}\right)\right)}^{2}\right)}^{- \frac{1}{2}} = {\left(\sin \left(2 - {x}^{2}\right)\right)}^{-} 1$
In order to differentiate this, we need to use the chain rule twice on $\sin {\left(f \left(x\right)\right)}^{n}$.
$\frac{d}{\mathrm{dx}} {\left[\sin \left(f \left(x\right)\right)\right]}^{n} = n {\left[\sin \left(f \left(x\right)\right)\right]}^{n - 1} \cos \left(f \left(x\right)\right) f ' \left(x\right)$
$f ' \left(x\right) = - 1 {\left({\sin}^{2} \left(2 - {x}^{2}\right)\right)}^{- 2} \cos \left(2 - {x}^{2}\right) - 2 x =$
$= \frac{2 x \cos \left(2 - {x}^{2}\right)}{{\sin}^{2} \left(2 - {x}^{2}\right)} ^ 2$