\(\frac{-5x}{(1-x^2)^6}\)
\(\frac{-10x}{(1-x^2)^6}\)
\(\frac{5x}{(1-x^2)^6}\)
\(\frac{10x}{(1-x^2)^6}\)
Correct answer is D
\(y = \frac{1} {(1 - x^2)^5} = (1-x^2)^{-5}\)
Let u = \(1 - x^2; y = u ^{-5}\)
\(\frac{du}{ dx}=-2x;\frac{dy}{ du}=-5u^{-6}\)
Using chain rule:
\(\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{ dx}\)
\(\frac{dy}{dx}=-2x\times-5u^{-6}\)
\(\frac{dy}{dx}=-2x\times-5(1 - x^2)^{-6}\)
\(\therefore\frac{dy}{dx} = 10x(1-x^2)^{-6}=\frac{10x}{(1-x^2)^6}\)