Using the chain rule to find the derivative of $f(x,y) = \sin(x\sin y)$.

I write, $\sin(x \sin y) = \sin \circ (x \sin y) = \sin \circ (p \circ (x,\sin y))$ where $p(x,y) = xy$. Now the chain rules gives

$$Df(x,y)(a,b) = D\sin (p \circ(x,\sin y) \circ D(p \circ(x, \sin y))(a,b)$$

$$ = D\sin (x \sin y) Dp(x,\sin y)D(x,\sin y))(a,b)$$ 

Now I use the reuslt that $Dp(x,y)(a,b) = ay + bx$.

$$Dp(x,\sin y)D(x,\sin y))(a,b) = Dp(x,\sin y)(a, \cos y b) = a \sin y + b x \cos y$$, so I am left to calculate $D\sin (x \sin y)$, which I am unable to do.