\(y = \frac{1} {(1 - x^2)^5} = (1-x^2)^{-5}\)

Let u = \(1 - x^2; y = u ^{-5}\)

\(\frac{du}{ dx}=-2x;\frac{dy}{ du}=-5u^{-6}\)

Using chain rule:

\(\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{ dx}\)

\(\frac{dy}{dx}=-2x\times-5u^{-6}\)

\(\frac{dy}{dx}=-2x\times-5(1 - x^2)^{-6}\)

\(\therefore\frac{dy}{dx} = 10x(1-x^2)^{-6}=\frac{10x}{(1-x^2)^6}\)