Letting $\Delta x\to0$ in multivariable chain rule

Question

In deriving the multivariable chain rule we can start with: $$\Delta u\approx \frac{\partial u}{\partial x} \Delta x +\frac{\partial u}{\partial y} \Delta y$$ and divide both sides by $\Delta t$ say to get: $$\frac{\Delta u}{\Delta t}\approx \frac{\partial u}{\partial x} \frac{\Delta x}{\Delta t}+\frac{\partial u}{\partial y} \frac{\Delta y}{\Delta t}$$ So here is my question: why when we let all the small quantities $\Delta a$ tend to $0$ do we get partial derivatives and not total derivatives. (I know that $u$ depends on other variables not just $t$ and am looking for a reason that is more well-defined.)

Answer 1

Let's approximate $u(x,y)$ near the point $u_0 = u(x_0,y_0)$. If you know how the tangent plane is constructed we have

$$u - u_0 \approx \frac{\partial u}{\partial x}(x_0,y_0) (x-x_0) + \frac{\partial u}{\partial y}(x_0,y_0) (y-y_0)$$

Now suppose that you parametrized $(x,y)$ as $(x(t),y(t))$ in such a way that $x_0 = x(t_0)$, $y_0 = y(t_0)$ and that the parametrization is differentiable. Then you can write the quotient

$$\frac{u - u_0}{t-t_0} \approx \frac{\partial u}{\partial x}(x_0,y_0) \frac{x-x_0}{t-t_0} + \frac{\partial u}{\partial y}(x_0,y_0) \frac{y-y_0}{t-t_0}$$

Letting $t \to 0$ gives

$$\frac{du}{dt}(x_0,y_0) = \frac{\partial u}{\partial x}(x_0,y_0) \frac{dx}{dt}(t_0) + \frac{\partial u}{\partial y}(x_0,y_0)\frac{dy}{dt}(t_0)$$

We have equality because the linear approximation becomes better as $(x,y) \to (x_0,y_0)$. Omitting the specific point we're doing this at yields the chain rule

$$\frac{du}{dt}= \frac{\partial u}{\partial x} \frac{dx}{dt}+ \frac{\partial u}{\partial y}\frac{dy}{dt}$$

The reason that you get $dx/dt$ and not $\partial x/ \partial t$ is because $x$ depends only on $t$ so the two quantities are the same. Now, the question may be where do the partial derivatives come from. For that you'd need to construct the tangent plane itself.