Answer
$\dfrac{x^2-2x}{(x-1)^2} \sin (\dfrac{x^2}{x-1})$
Work Step by Step
Let us suppose that $f(x)=\cos (\dfrac{x^2}{x-1})$
Use $\dfrac {d}{dx} (\cos a)=-\sin a \dfrac{da}{dx}$
We differentiate both sides with respect to $x$.
$f^{\prime}(x)=\dfrac{d}{dx} [\cos (\dfrac{x^2}{x-1})] \\=-\sin (\dfrac{x^2}{x-1})\times \dfrac{(x-1) \times(2 x)-x^2 }{(x-1)^2} $
Simplify to obtain:
$f^{\prime}(x)=\dfrac{x^2-2x}{(x-1)^2} \sin (\dfrac{x^2}{x-1})$
