Problem Set 2.5 - Chain Rule

 16. $ y= cos^3(\frac{x^2}{1 - x}) $

$ v= \frac{x^2}{1-x}, ~ u=cos \, v, ~ y = u^3 $

$\frac{dy}{dx} = \frac{dv}{dx}. \frac{du}{dv}. \frac{dy}{du}$

$ = (\frac{(1 - x)D_x(x^2) - (x^2) D_x(1 - x)}{(1 - x)^2}). (- sin \, v).(3u^2) $

$ = \frac{(1-x)(2x) -(x^2)(-1)}{(1 - x)^2}. sin(\frac{x^2}{1 -x}). (-3 cos^2 (\frac{x^2}{1 - x}))$

$ = \frac{-3(2x - x^2)}{(1 - x)^2}cos^2(\frac{x^2}{1-x}) sin(\frac{x^2}{1 -x }) $