# Total and directional derivatives:

1. May 25, 2012

### iampaul

The total derivative of the function z=f(x,y) with respect to x is:
dz/dx = ∂z/∂x + (∂z/∂y)(dy/dx)
The way i see this is that the total derivative, dz/dx, gives the rate of change of z with x, allowing y to vary with x at the rate dy/dx. I don't know if this is right.

The directional derivative Daf(x,y) gives the rate of change of z in the direction of the vector a. But, im thinking, that for a particular direction, there is a line: y=mx +b, and y is a function of x. A certain direction gives a unique dy / dx. If i substitute this in the equation for a total derivative, I get a unique dz/dx.

My question is, what really is a total derivative?

2. May 25, 2012

### HallsofIvy

It's pretty much just what you have said. The directional derivative, in a specific direction, tells how fast the function value changes as you move an "infinitesmal" distance in that direction per unit distance moved. The total derivative, with respect to x, in a given direction, measures how fast the function value changes as you move an "infinitesmal" distance in that direction per unit of change in x.

That is, for the directional derivative, the "denominator" is measured along the direction while for the total derivative it is measured in the x direction.

3. May 26, 2012

### algebrat

Um, but In some cases, total derivative is the jacobian, so the total derivative is actually the gradient in this case. Then once you restrict your self to a direction, you have a new function just in x, and now that one dimesnional derivtaive could be seen as a directional derivative. One problem is, when people write f(x)=f(x,y(x)), that's not right. It should have a new name, g(x)=f(x,y(x)). Total derivative of f is gradient. total derivative of g id dg/dx=f_x+f_y*y'=grad f dotted with u, the unit tangent vector to the y(x) path. In other words, sometimes the difference between what your teachers call partial and total, is fixing a path.

4. May 26, 2012

### iampaul

Thanks for the replies. I know nothing about those jacobians yet. I am using the book protter morrey college calculus which i got from my uncle, and it mentioned nothing about this total derivatives, so i easily got confused. I'll post a reply when i run into other problems. Thanks again.

5. May 26, 2012

### algebrat

Let me try to paraphrase my messy statement.

I think we can imagine the words partial and total, as being used somewhat appropriately.

• The partial moves us just along the x-axis, and everything else is held constant.
• The total derivative appears to have more or less two meanings, though in my own head, I see them as the same:
1. The total derivative is a derivative from multivariable calculus which records all the partials at once, in a list, but also in an abbreviated notation. This is the Jacobian, and in a special case the gradient; wikipedia suggests it is the same from differential forms for manifolds, sounds about right.
2. For the other meaning, I feel there is still the spirit of the total derivative recording rates of changes in every direction, but in order to only have one value, we must be looking at one particular direction. So in the case of restricting ourselves to some path, the total derivative is not a partial, and is not multivariable. It is slightly different from the directional derivative, but it's similar in that that is the direction in which we want to measure the change of f. But the speed at which we travel horizontally is different, since dx is at an angle from a direction du in which we measure fs change. I think the total derivative will be different from the directional derivtiave along same path by a factor $\cos(/theta)$, where theta is the angle between the direction u and the positive x axis. Maybe this fails for a couple angles.

Sorry for the rambling. Halls of ivy's answer was great, I just only rememebered total from another class.

6. May 27, 2012

### iampaul

7. May 27, 2012

### iampaul

I have another question about an equation I arrived at while attempting to clarify my confusion.
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Daf(x,y) = limh→0{[f(x+hcos∅,y+hsin∅)-f(x,y)]/h}. I think that h is always given by Δy/sin∅ or Δx/cos∅.When i plugged these into the equation for the directional derivative I get:
Daf(x,y) = limh→0{[f(x+Δx,y+Δy)-f(x,y)]/Δy}sin∅

or

Daf(x,y) = limh→0{[f(x+Δx,y+Δy)-f(x,y)]/Δx}cos∅

when h→0, Δy→0 and Δx→0

Daf(x,y)= (df/dx)cos∅ = (df/dy)sin∅

Is this right?? Why is it that when ∅=0 i get Daf(x,y)=df/dx=0..The same thing happens when ∅= ∏/2,∏,3∏/2. Is it because h is undefined in these angles: h=Δy/sin∅=Δx/cos∅?
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I get the same result if i start from:

dz/dx = (∂f/∂x) + (∂f/∂y)dy/dx......dy/dx = tan∅ for a certain direction
(dz/dx)cos∅=(∂f/∂x)cos∅+(∂f/∂y)sin∅=Daf(x,y)

the same thing happens with dz/dy