How to find $$\frac{\partial^2u}{\partial x \partial y}$$ given $$u=\sin(x\sin^{-1}(y))$$?
I have calculated
$$\frac{\partial u}{\partial x }=\sin^{-1}(y)\cdot \cos(x\sin^{-1}(y))$$ but get stuck on applying the product rule on the next derivative namely finding the partial of $\cos(x\sin^{-1}(y))$ w.r.t $y$.
Hint: Let $$z=\sin^{-1}y\implies \partial^2_{xy}u=\partial_y(z\cos(xz))=\frac{dz}{dy}(\cos(xz)-xz\sin (xz)$$ I think you can proceed from here.
$u=sin(x \arcsin(y))$
$\frac{\partial u}{\partial y}=cos(x \arcsin(y)) \times x\times\frac{1}{\sqrt(1-y^2)}$ (treat x as constant)
$\frac{\partial^2 u}{\partial x \partial y}=\frac{1}{\sqrt(1-y^2)}\big[-x\sin(x \arcsin(y))\times \arcsin(y)+cos(x \arcsin(y)\big]$ (treat y as constant)
Let $f(x,y)=\sin (x\arcsin (y))$. The partial derivative with respect to $x$ is
$$\frac{\partial f(x,y)}{\partial x}=\cos (x\arcsin (y))\times \arcsin (y)$$
The mixed partial with respect to $x$ and $y$ is then
$$\begin{align} \frac{\partial^2 f(x,y)}{\partial x\partial y}&=\frac{\partial }{\partial y}\left(\cos (x\arcsin (y))\times \arcsin (y)\right)\\\\ &=\cos (x\arcsin (y))\times \frac{1}{\sqrt{1-y^2}}-\arcsin (y)\times \sin (x\arcsin (y))\times \frac{x}{\sqrt{1-y^2}} \end{align}$$
