y=1(1−x2)5=(1−x2)−5y = \frac{1}{(1 - x^2)^5} = (1-x^2)^{-5}y=(1−x2)51=(1−x2)−5
Let u = 1−x2; y=u−51 - x^2;\ y = u^{-5}1−x2; y=u−5
dudx=−2x; dydu=−5u−6\frac{du}{dx} = -2x;\ \frac{dy}{du} = -5u^{-6}dxdu=−2x; dudy=−5u−6
Using chain rule:
dydx=dydu×dudx\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}dxdy=dudy×dxdu
dydx=−2x×−5u−6\frac{dy}{dx} = -2x \times -5u^{-6}dxdy=−2x×−5u−6
dydx=−2x×−5(1−x2)−6\frac{dy}{dx} = -2x \times -5(1 - x^2)^{-6}dxdy=−2x×−5(1−x2)−6
∴dydx=10x(1−x2)−6=10x(1−x2)6\therefore \frac{dy}{dx} = 10x(1-x^2)^{-6} = \frac{10x}{(1-x^2)^6}∴dxdy=10x(1−x2)−6=(1−x2)610x
