# Partial Derivatives - Basic Formula

1. Nov 23, 2012

### elemis

Could someone please explain how the formula at the bottom of the page is derived i.e. how is the Taylor theorem used to obtain it ?

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2. Nov 23, 2012

### arildno

You can derive it, by expanding one variable by one.

f(h+x_0, k+y_0)=f(x_0,y_0+k)+f_x(x_0,y_0+k)h+1/2f_xx(x_0,y_0+k)h^2+++
=f(x_0,y_0)+f_y(x_0,y_0)k+1/2f_yy(x_0,y_0)k^2+f_xy(x_0,y_0)hk+1/2f_xx(x_0,y_0)h^2+++

Where the terms up to second order can be rearranged into the desired equation.

3. Nov 23, 2012

### elemis

I'm not sure what you've done there to be honest.

4. Nov 23, 2012

### arildno

First, I make a one-variable Taylor-expansion of the function in x, at the point (x_0,y_0+k).
Then, I make a one-variable expansion in y of the different terms.

5. Nov 24, 2012

### elemis

I have to be truthful, my professor used a similar method to what you described but I just cant see how he used two variables in a Taylor expansion since I've only ever learnt to deal with Taylor expansions of one variable.

Could you please break this down further for me ? I would be very much obliged.

6. Nov 24, 2012

### arildno

Well, but when you do partial derivatives, you regard the other variable to be a CONSTANT.
Agreed?

Thus, the function f(x,y), evaluated at the 2-D point (x_0+h,y_0+k) may be regarded as a single-variable function of x along the (horizontal) length segment from (x_0, y_0+k) to (x_0+h, y_0+k).
Along THAT line segment, we may perfectly well expand the function in its Taylor series relativ to "x".

7. Nov 24, 2012

### lurflurf

In one variable we have

f(x+h)~exp(hD)f(x)=f(x)+hDf(x)+(h^2)(D^2)/2f(x)+...

in several variables

f(x+h)~exp(h.∇)f(x)=f(x)+h.∇f(x)+(h.∇)^2/2f''(x)+...

8. Nov 24, 2012

### arildno

Why, lurfflurff, do you come up with compact notation a newbie couldn't possibly understand??

9. Nov 24, 2012

### elemis

Yes, I agree and understand what you mean.

The difficulty I'm having is how do I introduce the constant y_0+k into my taylor expansion ?

10. Nov 24, 2012

### arildno

At whatever places "y" stands in the expression for "f" itself.

Furthermore, once you have expanded "f" along the horizontal length segment from (x_0, y_0+k) to (x_0+h, y_0+k), you may expand the different terms of "f" along the VERTICAL length segment from (x_0,y_0) to (x_0, y_0+k).