Total and partial derivatives

Question

I have a function $Y(x,t)$ and another function $\eta(x,t)$.

I have the total derivative $\dfrac{dY}{d\eta}$.

So, what is the correct?

$$\dfrac{dY}{d\eta}\dfrac{\partial \eta}{\partial x}=\dfrac{\partial Y}{\partial x}\dfrac{\partial x}{\partial \eta}\dfrac{\partial \eta}{\partial x}+\dfrac{\partial Y}{\partial t}\dfrac{\partial t}{\partial \eta}\dfrac{\partial \eta}{\partial x}=\dfrac{2\partial Y}{\partial x}$$

or

$$\dfrac{dY}{d\eta}\dfrac{\partial \eta}{\partial x}=\dfrac{\partial Y}{\partial x}\dfrac{\partial x}{\partial \eta}\dfrac{\partial \eta}{\partial x}+\dfrac{\partial Y}{\partial t}\dfrac{\partial t}{\partial \eta}\dfrac{\partial \eta}{\partial x}=\dfrac{\partial Y}{\partial x}+\dfrac{\partial Y}{\partial t}\underbrace{\dfrac{\partial t}{\partial x}}_{0}=\dfrac{\partial Y}{\partial x}$$

?

Thank you so much.

Answer 1

$$\frac{\mathrm{d}Y}{\mathrm{d}\eta}=\frac{\partial Y}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}\eta}+\frac{\partial Y}{\partial t}\frac{\mathrm{d}t}{\mathrm{d}\eta}$$ Now, $$\frac{\mathrm{d}\eta}{\mathrm{d}x}=\frac{\partial \eta}{\partial x}+\frac{\partial \eta}{\partial t}\frac{\mathrm{d}t}{\mathrm{d}x}$$ $$\frac{\mathrm{d}\eta}{\mathrm{d}t}=\frac{\partial \eta}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}t}+\frac{\partial \eta}{\partial t}$$ Invert these and plug them into the above. This all tends to be much easier when we have some information about $Y,\eta$.

Answer 2

In the described situation there is no such thing as ${dY\over d\eta}$.

When $$Y: \quad(x,t)\mapsto Y(x,t),\qquad \eta:\quad (x,t)\mapsto\eta(x,t)$$ are two real valued functions defined in the $(x,t)$-plane, and a point $(x_0,t_0)$ is given, then the gradients of $Y$ and $\eta$ at $(x_0,t_0)$ determine the increments of $Y$ and $\eta$ when moving away from $(x_0,t_0)$: $$\eqalign{\Delta Y&=\nabla Y(x_0,t_0)\cdot(\Delta x,\Delta t)+o\bigl(\sqrt{\Delta x^2+\Delta t^2}\bigr),\cr \Delta \eta&=\nabla \eta(x_0,t_0)\cdot(\Delta x,\Delta t)+o\bigl(\sqrt{\Delta x^2+\Delta t^2}\bigr)\ .\cr}$$ It follows that $\Delta Y$ and $\Delta\eta$ depend on the direction of the increment vector $(\Delta x,\Delta t)$. The same is true for the quotient ${\Delta Y\over\Delta\eta}$ and its limit (called ${dY\over d\eta}$ by you), when $\sqrt{\Delta x^2+\Delta t^2}\to0$.

It is another thing when $x$ and $t$ are not independent variables, and $x$ is in fact a function of $t$, given as $x=\psi(t)$. In this case $Y$ and $\eta$ also become composed functions of $t$: $$\hat Y(t):=Y\bigl(\psi(t),t\bigr),\qquad \hat \eta(t):=\eta\bigl(\psi(t),t\bigr)\ .$$ In this case it makes sense to talk about $${d\hat Y\over d\hat\eta}={\hat Y'(t)\over\hat\eta'(t)}=\ldots\quad .$$