How can we prove rule of derivative of multivariable functions.

Question

I am waiting for all methods but especially define of limit of derivative.

Attemt:

Let $x$ and $y$ depend on $t$,

and we define $f(x(t),y(t))$ function that is differentiable.

$$\dfrac{df}{dt}=\dfrac{\partial f}{\partial x}\dfrac{dx}{dt}+\dfrac{\partial f}{\partial y}\dfrac{dy}{dt}$$

That is;

If we take $U(x_1(t),x_2(t),....,x_i(t))$ $$\dfrac{dU}{dt}=\dfrac{\partial U}{\partial x_1}\dfrac{dx_1}{dt}+\dfrac{\partial U}{\partial x_2}\dfrac{dx_2}{dt}+...+\dfrac{\partial U}{\partial x_i}\dfrac{dx_i}{dt}$$

But why?

To prove, I tried to apply definition limit of derivative;

Let define, $G(t)=f(x(t),y(t))$

$$\dfrac{dG}{dt}=\lim\limits_{h\to 0}\dfrac{G(t+h)-G(t)}{h}=\lim\limits_{h\to 0}\dfrac{f(x(t+h),y(t+h))-f(x(t),y(t))}{h}$$

And let's add and remove $f(x(t+h),y(t))$ and $f(x(t),y(t+h))$

$$\dfrac{dG}{dt}=\lim\limits_{h\to 0}\dfrac{2G(t+h)\pm f(x(t+h),y(t))-\pm f(x(t),y(t+h))-2G(t)}{2h}$$

$$\dfrac{dG}{dt}=A+B+C+D$$

,So they are;

$$A=\lim\limits_{h\to 0}\dfrac{f(x(t+h),y(t+h))-f(x(t+h),y(t))}{2h}$$

$$B=\lim\limits_{h\to 0}\dfrac{f(x(t+h),y(t+h))-f(x(t),y(t+h))}{2h}$$

$$C=\lim\limits_{h\to 0}\dfrac{f(x(t),y(t+h))-f(x(t),y(t))}{2h}=\dfrac12\dfrac{\partial f}{\partial y}$$

$$D=\lim\limits_{h\to 0}\dfrac{f(x(t+h),y(t))-f(x(t),y(t))}{2h}=\dfrac12\dfrac{\partial f}{\partial x}$$

$C$ and $D$ O.K. but I didnt exactly understand $A$ and $B$ and couldn't get given formula also I want to understand the physical meaning of this.

Answer 1

This is not the way I would have done it, though the following is not at all rigorously developed. Now, if we try to apply the limit definition, (as $h\to0$)

$$\begin{align}\frac{dG}{dt}&=\frac{f(x(t+h),y(t+h))-f(x(t),y(t))}h\\&=\frac{f(x(t+h),y(t+h))\color{#3388dd}{-f(x(t+h),y(t))+f(x(t+h),y(t))}-f(x(t),y(t))}h\\&=\frac{f(x(t+h),y(t+h))-f(x(t+h),y(t))}h+\frac{f(x(t+h),y(t))-f(x(t),y(t))}h\end{align}$$

We'll call the first fraction $A$ and the second fraction $B$.

$$\begin{align}A&=\frac{f(x(t+h),y(t+h))-f(x(t+h),y(t))}h\\&=\frac{f(x(t+h),y(t+h))-f(x(t+h),y(t))}{\color{#3388dd}{y(t+h)-y(t)}}\frac{\color{#3388dd}{y(t+h)-y(t)}}h\end{align}$$

Now, if this limit exists, then,

$$A=\lim_{h_x\to0}\lim_{h\to0}\frac{f(x(t+h_x),y(t+h))-f(x(t+h_x),y(t))}{y(t+h)-y(t)}\frac{y(t+h)-y(t)}h\\=\lim_{h_x\to0}\frac{\partial f(x(t+h_x),y)}{\partial y}\frac{dy}{dt}\\=\frac{\partial f}{\partial y}\frac{dy}{dt}$$

Similarly, you will find that $B=\frac{\partial f}{\partial x}\frac{dx}{dt}$, though with no requirement of splitting the limit like we did above..

Answer 2

Let $x_i: A \subset \Bbb R \to \Bbb R$ be differentiable on $A$ and $U: B \subset \Bbb R^n \to \Bbb R$ be differentiable on $B$, where $B \supset A\times\cdots \times A$. Let $G(t) = U(x_1(t),\ldots, x_n(t))$, i.e. $G = U \circ(x_1,\ldots,x_n)$, we have by the chain rule:

$$\frac{dG}{dt} = \nabla U(x_1,\cdots, x_n) \cdot \left(\frac{dx_1}{dt}, \ldots, \frac{dx_n}{dt}\right) = \left(\frac{\partial U}{\partial x_1}, \ldots, \frac{\partial U}{\partial x_n}\right) \cdot \left(\frac{dx_1}{dt}, \ldots, \frac{dx_n}{dt}\right) \\ = \sum_{i=1}^n \frac{\partial U}{\partial x_i} \frac{dx_i}{dt}$$

Just to be clear:

• $\frac{d(\cdot)}{dt} = (\cdot)'(t)$.

• $\nabla U(x_1,\ldots,x_n) = \nabla U(x_1(t),\ldots,x_n(t))$.

• $\frac{\partial U}{\partial x_i} = \frac{\partial U}{\partial x_i}(t)$

I hope this helps.

Answer 3

I will try to explain the formula through its geometrical meaning.

(1) $z=G(x,y)$ describes a surface in the $x,y,z$ coordinate system.

(2) $(x(t),y(t))$ describes a curve in the $x,y$ plane.

Let, for a $t_0$: $x(t_0)=x_0$ and $y(t_0)=y_0$

If $t$, the parameter of the curve changes from $t_0$ to $t_0+\Delta t$ then $G$ changes and let the total change of $G$, $\Delta G$ be defined as

$$\Delta_x G+\Delta_y G.\tag 1$$

where $$\Delta_xG=G(x(t_0+\Delta t),y_0)-G(x(t_0),y_0)$$ and $$\Delta_yG=G(x_0,y(t_0+\Delta t))-G(x(t_0),y_0).$$

Then

$$\frac{\Delta G}{\Delta t}=\frac{\Delta_xG+\Delta_xG}{\Delta t}=$$ $$=\frac{G(x(t_0+\Delta t),y_0)-G(x(t_0),y_0)}{\Delta t}+\frac{G(x_0,y(t_0+\Delta t))-G(x(t_0),y_0)}{\Delta t}.$$

Letting $\Delta t\to 0$, we get by the chain rule (if these limits exist)

$$\frac{d G}{dt}=\frac{d G(x(t),y_0)}{dt}+\frac{d G(x_0,y(t))}{dt}=$$ $$=\frac{\partial G(x,y_0)}{\partial x}\frac{d x}{d t}+\frac{\partial G(x_0,y)}{\partial y}\frac{d y}{d t}.$$

A simpler way of writing the same is

$$\frac{d G}{dt}=\frac{\partial G}{\partial x}\frac{d x}{d t}+\frac{\partial G}{\partial y}\frac{d y}{d t}.\tag 2$$

Or

$$\frac{d G}{dt}=\nabla G \cdot v$$

where $\nabla G=\left(\frac{\partial G}{\partial x},\frac{\partial G}{\partial y} \right)$ is the gradient vector field of $G$ and $v=\left(\frac{d x}{d t},\frac{d y}{d t}\right)$ is the tangent vector to the curve defined by $(x(t),y(t))$. So, $\frac{d G}{dt}$ is the directional derivative of $G$ in the direction $v$.

---

The definition of the "total change of $G$" by $(1)$ may seem to be arbitrary. But the result justifies that choice.

For example, if $r=(x(t),y(t))$ then $v=\dot r$ the velocity of a particle moving along the curve at stake. Then $\frac{d G}{dt}$ is the rate of change of the height (with respect to the time) measured by a traveler skating on the surface $G$ always right above $r$.

$(1)$ and $(2)$ can be generalized for higher dimensions. Then the traveler's example can be applied again.

Answer 4

Suppose $(a,b) \in U,$ where $U$ is open in $\mathbb R^2.$ Assume $f:U\to \mathbb R,$ and $f$ is differentiable at $(a,b).$ This means, by definition, that there is a linear transformation $T:\mathbb R^2 \to \mathbb R$ such that

$$f(x,y) - f(a,b) = T(x-a,y-b) + ((x-a)^2+(y-b)^2)^{1/2}\,r(x,y),$$

where $\lim_{(x,y)\to (a,b)}r(x,y)=0.$

Now $T$ has the form $T(x,y) = Ax + By.$ And the constants $A,B$ are what you would expect: $A= \partial f /\partial x (a,b),B= \partial f /\partial y (a,b).$

Now assume $g(t) = (x(t),y(t))$ maps a neighborhood of $t_0\in \mathbb R$ into $U,$ with $x(t_0)=a, y(t_0)=b.$ Assume $g'(t_0) = (x'(t_0),y'(t_0))$ exists. Then

Thm: $(f\circ g)'(t_0) = \nabla f (g(t_0))\cdot g'(t_0),$ where $\cdot $ denotes the dot product.

Proof: We're all set up:

$$\frac{f\circ g(t)-f\circ g(t_0)}{t-t_0} = A\frac{x(t)-x(t_0)}{t-t_0} + B\frac{y(t)-y(t_0)}{t-t_0}$$ $$ + \frac{((x(t)-x(t_0))^2+(y(t)-y(t_0))^2)^{1/2}}{t-t_0}r(x(t),y(t)).$$

As $t\to t_0,$ the first summand on the right $\to A x'(t_0),$ and the second summand $\to By'(t_0).$ As for the third summand, verify that the fraction is bounded, and that $r(x(t),y(t))\to 0.$ This shows the third summand $\to 0$ and we have the result.