$Explain partial derivatives v3$

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$Explain partial derivatives v3$

This time I want to discuss the proof(part of it) of $$D_uf(x,y)=f_x(x,y)a + f_y(x,y)b, u = <a,b>$$
if we define a function g of the single variable h:$$g(h)=f(x_0+ha,y_0+hb)$$then, by definition of derivative, we have$$g'(0)=\lim_{{h \to 0}} \frac{g(h)-g(0)}{h} = \lim_{{h \to 0}} \frac{f(x_0+ha,y_0+hb) - f(x_0,y_0)}{h}$$
So what I understand :
If I have $$P(x_0,y_0,0) and Q(x_0+ha,y_0+hb,0) and u = <a,b>$$h represents the length of PQ(right?).
So it's clear why $$\lim_{{h \to 0}} \frac{f(x_0+ha,y_0+hb) - f(x_0,y_0)}{h}$$
But what I don't understand is when we define a function g of the single variable h, h represents the length of the vector(right?), but when we are trying to get the derivative of g(any derivative for h), what I will get is $$\frac{dg}{dh} = \lim_{{h \to 0}} \frac{g(h + h) - g(h)}{h}$$
So it makes no sense, since h is ambiguous, and it means that those h's represent different values? It seems to be an easy question, but i've been studying for whole day and probably not understanding something.
P.S. let me know If you feel like an answer worth more money

Calculus

Babaduras

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Mathe

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I don't fully understand your question. And I think that you meant g(x+h)-g(x) in your last equation.
- Babaduras
  
  0
  
  but g is function of h, not x, that's exactly what I don't understand either
Babaduras

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i mean that when calculating g'(0) it equals to (lets skip limit) (g(h)-g(0))/h, so for the function g(h) in it's derivative h represents the difference of independent variable(which approaching 0), but for the same time g is function of h, so I'm little bit confused

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Mathe

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I don't fully understand your question. And I think that you meant g(x+h)-g(x) in your last equation.

Babaduras

0

but g is function of h, not x, that's exactly what I don't understand either
Babaduras

0

but g is function of h, not x, that's exactly what I don't understand either
Babaduras

0

i mean that when calculating g'(0) it equals to (lets skip limit) (g(h)-g(0))/h, so for the function g(h) in it's derivative h represents the difference of independent variable(which approaching 0), but for the same time g is function of h, so I'm little bit confused

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I think I understand where the issue is and hence your confusion.

It seems like you're mixing up two different contexts involving the variable 'h': one related to the derivative of a function of a single variable, and the other related to the partial derivative of a multivariable function.

Case 1, g(h):

When you're dealing with a function of a single variable, i.e. g(h), the variable 'h' represents the input of the function. In this case, 'h' is just a scalar value, representing a distance along a one-dimensional line. The derivative of g with respect to 'h', denoted as dg/dh, represents how the function g changes as you change the input 'h'. So, the expression:

$$\frac{dg}{dh}=\lim_{h \to 0} \frac{g(x+h)-g(x)}{h}$$

means that you're looking at how the function g changes as you make a small change in the input 'h'. In this context, 'h' is not a vector; it's a scalar representing the distance along a one-dimensional line.

Case 2, f(x, y):

In the partial derivative context, 'h' is being used to represent a vector in the two-dimensional plane. Specifically, if you have a multivariable function $f(x, y)$, and you want to find the partial derivatives with respect to 'x' and 'y', you introduce a vector $u = <a, b>,$ which represents the change you want to make in the x and y directions. Here, 'u' is a vector, and 'h' (which you're calling the length of 'u') represents the magnitude of that vector. 'h' is used to control the magnitude of the change you're making in the x and y directions.

Now, regarding the equation you mentioned:

$$D_{u}f(x,y)=f_{x}(x,y)a + f_{y}(x,y)b$$

This equation expresses the directional derivative of the function $f(x, y)$ in the direction of the vector $u = <a, b>$. It's saying that the change in the function $f(x, y)$ as you move in the direction of u can be approximated by the sum of two components: one in the x-direction (a times the partial derivative with respect to x) and one in the y-direction (b times the partial derivative with respect to y). This is a key concept in multivariable calculus.

So, make sure you're clear about the context in which 'h' is being used: as a scalar in the derivative of a single variable function OR as the magnitude of a vector in the context of the partial derivative of a multivariable function. These are distinct concepts, and understanding their differences should help alleviate your confusion.

This took me about 30 minutes, my rate is $50/hour, so it is worth more than the bounty. Thanks.

Kav10

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Babaduras

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I don't call 'h' the length of 'u' since 'u' is a unit vector, it's length is 1. I'm calling 'h' the length of projection vector, which appears when I want to get the derivative at particular point, and choose other point on some distance 'h' from it in the direction of unit vector.
- Kav10
  
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  Yes. g(h) represents the value of f at a point that is h units away from the initial point (x_0, y_0) in the direction of u. So, h is indeed the distance of the "projection vector" along the direction of u. The derivative g'(0) represents the rate of change of g with respect to h when h is very close to 0. This is exactly what you need to find the directional derivative of f in the direction of u.
- Kav10
  
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  So, D_uf(x, y) gives you the rate of change of f in the direction of the unit vector u at the point (x,y), and it's approximated by combining the partial derivatives of f with respect to x and y along the direction of u.
Babaduras

0

@Kav10 yeah, I understand the logic behind it, finding derivative when distance is 0 equals to the actual rate of change. But what I still didn't get is why when I differentiate g(h) h represents the length of the projection vector, to find the derivative I have to find a limit of the distance for (g(independent var + distance) - g(independent var))/distance, so even if I receive (omit the limit) g(x + h) - g(x)/h, what does 'h' represent here, when x represents independent variable of g?
- Kav10
  
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  I think there might be a slight misunderstanding about the notation and its interpretation. Let me try to clarify again.
- Kav10
  
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  You define g(h) as a function that represents the value of the multivariable function f(x,y) at a point that is a distance h away from the initial point (x_0, y_0) in the direction of the unit vector u=⟨a,b⟩. This can be written as g(h) = f(x_0 + ha, y_0 + hb). Here, h represents the distance along the direction of u, and x_0 + ha and y_0 + hb represent the coordinates of the point at that distance.
- Kav10
  
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  When you calculate the derivative g'(h), you are looking at how the function g(h) changes as you change the distance h. The derivative is defined as g'(h) = \lim_{{\Delta h \to 0}} \frac{g(h + \Delta h) - g(h)}{\Delta h} The notation h+Delta h means you're considering a point that is a distance h+Delta h away from the initial point.
- Kav10
  
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  The expression g \frac{g(h + \Delta h) - g(h)}{\Delta h measures the rate of change of g(h) with respect to h, as you vary the distance h. In this context, h is not a variable representing the independent variable of g(h), but rather the distance along the direction of u.
- Kav10
  
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  The derivative g'(h) measures how g(h) changes as you move along the direction of u by changing the distance h. Remember that the derivative g'(h) provides information about how g(h) changes as you vary h, which represents the distance along the direction defined by the unit vector u.
- Babaduras
  
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  LaTeX doesn't work in comments section :)
- Kav10
  
  +1
  
  If you're looking for the rate of change of the function f(x,y) in the direction of the unit vector u at a specific point (x_0, y_0), then you are indeed interested in the derivative D_uf(x_0, y_0), and the expression g(h) and its derivative g'(h) are tools used to help understand this concept.
- Kav10
  
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  I hope this explanation clarifies things!
Babaduras

0

that is clear for me.

$Explain partial derivatives v3$

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