Partial derivatives of $u = \frac{\cos x^2}{y}$ (1st and 2nd)

Question

So there's

$$u = \frac{\cos(x^2)}{y}$$

I got the following:

$$\frac{\partial{u}}{\partial{x}} = -\frac{2x\sin(x^2)}{y}$$

$$\frac{\partial{u}}{\partial{}y} = -\frac{\cos(x^2)}{y^2}$$

$$\frac{\partial^2{u}}{\partial{y^2}} = 2\frac{\cos(x^2)}{y^3}$$

I am a bit confused with calculus of $\frac{\partial^2{u}}{\partial x^2}$

Answer 1

Well, \begin{align} \frac{\partial^2 u}{\partial x^2} &= \frac{\partial}{\partial x}\frac{\partial u}{\partial x} \\ &= \frac{\partial}{\partial x}\left(\frac{-2x \sin x^2}{y} \right) \\ &= -\frac{2}{y}\frac{\partial}{\partial x}\left(x \sin x^2\right) \end{align} Now the problem becomes an exercise in the chain rule.

Answer 2

Since it is independent, $y$ is constant when computing $\frac{\partial^2 u}{\partial x^2}$. Thus $$ \begin{align} \frac{\partial^2 u}{\partial x^2} &= -\frac{2}{y} \frac{\partial}{\partial x}(x \sin(x^2))\\ &= -\frac{2}{y} (\sin(x^2) + 2x\cos(x^2)). \end{align} $$