# Why df=(∂f/∂x)dx + (∂f/∂y)dy?

1. Oct 14, 2011

### ck00

let f(x,y)=0
Why df=(∂f/∂x)dx + (∂f/∂y)dy?

2. Oct 14, 2011

### Bacle2

This is the definition of the total derivative, aka differential as I know it.

df here gives you the equation of the tangent plane that approximates the change of the function near a point. Was that your question?

3. Oct 14, 2011

### HallsofIvy

If x and y are themselves functions of a parameter, say, t, then
we can think of f(x, y)= f(x(t), y(t)) as a function of the single variable t and, by the chain rule:
$$\frac{df}{dt}= \frac{\partial f}{\partial x}\frac{dx}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt}$$

And then, the usual definition of the "differential" as df= (df/dt)dt gives
$$df= \frac{\partial f}{\partial x}\frac{dx}{dt}dt+ \frac{\partial f}{\partial y}\frac{dy}{dt}dt= \frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial y} dy$$

Last edited by a moderator: Oct 15, 2011
4. Oct 14, 2011

### ck00

But how can I prove this $$\frac{df}{dt}= \frac{\partial f}{\partial x}\frac{dx}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt}$$?

I know the basic operation in partial differentiation but i just not quite understand the theory behind it. Are there some proofs?

5. Oct 15, 2011

### 3.1415926535

A very informal (and possibly incorrect) proof I just thought of:

$$df=df(x(t),y(t))=f(x(t+h),y(t+h))-f(x(t),y(t))=f(x(t+h),y(t+h))-f(x(t),y(t+h))+f(x(t),y(t+h))-f(x(t),y(t))=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy\Leftrightarrow \frac{df}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}$$

For something more formal http://math.uc.edu/~halpern/Calc.4/Handouts/Proofchainrule2dim.pdf

Last edited by a moderator: Apr 26, 2017
6. Oct 16, 2011

### Tosh5457

It's the chain rule, it's very used in calculus. You can see a demonstration in wikipedia.

That can be deduced writing f(x,y) as Taylor's series (for multivariate functions), and going up to the 2nd term. To do it f only has to be differentiable 2 times in (a,b) neighbourhood.

$$f(x,y) = f(a,b) + (\frac{\partial f}{\partial x}(a,b), \frac{\partial f}{\partial y}(a,b))\cdot (x-a, y-b)$$

Putting,
$$x-a=\Delta x$$
$$y-b=\Delta y$$
$$f(x,y)-f(a,b)=\Delta f$$

When $$\Delta x \to 0$$ and $$\Delta y \to 0$$ you get that expression.

7. Oct 16, 2011

### Bacle2

Now put the two previous answers together (they did the hard work, I am just chiming-in), and do a δ-ε proof, showing that you can approximate the value of your function within ε>0 by using the right value of δ. This is for real-valued functions. If not, i.e., for Rn-valued maps , show:

i)The differential df is a linear map:

ii) ||f(x+h)-f(x)-hL(x)||/||h||→ 0

as ||h||→0 is satisfied only by the differential L(x)=df

In your case, you want to show that your function can be approximated to any degree of accuracy ε>0 by working within a ball B(x,δ), as all the other posters said.