Confusion with partial derivatives

Question

Let $y=e^{t^2/2}$, so $\frac{dy}{dt}=yt$. Then $\frac{\partial^2y}{\partial t\partial y}=t$, although this would contradict $\frac{\partial^2y}{\partial y\partial t}=\frac{\partial 1}{\partial t}=0$ (and also the actual value $t+\frac1t$)

Where did I go wrong and what's the proper way to take the partial derivative with respect to $y$?

Answer 1

Note that $$y'=y(t)t$$

thus

$$\frac{\partial y'}{\partial y}=t+y\frac{d t}{d y}=t+\frac1t$$

I'm not completely sure that take this kind of partial derivative has a meaning but note that

$$\frac{\partial^2y}{\partial t\partial y}=\frac{\partial}{\partial t}\frac{\partial y}{\partial y}=\frac{\partial}{\partial t}1=0$$

$$\frac{\partial^2y}{\partial t\partial y}=\frac{\partial}{\partial y}\frac{\partial y}{\partial t}=\frac{\partial}{\partial y}te^{t^2/2}=0$$

Note also that

$$\frac{\partial y'(y,t)}{\partial y}=\frac{\partial}{\partial y}\frac{dy(t)}{dt}\neq\frac{\partial^2y(t)}{\partial t\partial y}=\frac{\partial}{\partial t}\frac{\partial y(t)}{\partial y}$$

since the LHS is a first order partial derivative on $y'(y,t)$ while the RHS is a second order partial derivative on $y(t)$.

Answer 2

If $y = e^{t^2/2}$ ($y$ function), $\frac{\partial y}{\partial y}$ does not makes sense because you are using the same name for different things.

$\frac{\partial(e^{t^2/2})}{\partial y}$ ($y$ variable) makes sense and is $= 0$.

$\frac{\partial y}{\partial y}$ ($y$ variable) makes sense and is $= 1$.