A sufficient condition for differentiability for many variables
From a graphical perspective, when the derivative of a single variable function is continuous at a point, we can safely assume that the function is differentiable there. This is a mathematical way of defining the "smoothness" of a function. The same concept can be extended to multivariable functions. If the partial derivatives are continuous, the function is differentiable. This theory is a shortcut, a tool, to be used when we want to find whether a function is differentiable or not without the definition.
One may have asked about the directional derivative. Remember, the directional derivative finds a rate of change with a certain direction, not a function. It's meaningless to study continuity of values. We study continuity of functions.
Careful with some assumptions here! From a previous theorem we know that being differentiable implies in being continuous, because if the function is not continuous it cannot be differentiable. However, certain continuous functions fail to be differentiable. In addition, the partial derivatives themselves are not sufficient to prove that the function is differentiable. The fact that the function is differentiable does not imply that the partial derivatives are continuous. The function must be continuous to be differentiable, but the derivatives themselves may be discontinuous.
Let [math]\displaystyle{ f }[/math] be a function with continuous partial derivatives in an open domain [math]\displaystyle{ D }[/math]. Then [math]\displaystyle{ f }[/math] is differentiable in all points in [math]\displaystyle{ D }[/math].
With the domain being open, there is an open ball with its center at [math]\displaystyle{ (x_0,y_0) }[/math], in [math]\displaystyle{ D }[/math]. Let [math]\displaystyle{ h }[/math] and [math]\displaystyle{ k }[/math] be such that [math]\displaystyle{ (x_0 + h,y_0 + k) }[/math] is in the ball. We have:
[math]\displaystyle{ f(x_0 + h, y_0 + k) - f(x_0, y_0) = \underbrace{f(x_0 + h, y_0 + k) - f(x_0,y_0 + k)}_{\text{I}} + \underbrace{f(x_0,y_0 + k) - f(x_0,y_0)}_{\text{II}} }[/math]
Look at the above graph and locate the points. It's pretty clear the location of each one. We are assuming the function to be differentiable, which implies in it being continuous. If you follow the coordinates of each point, there are no gaps inside the circle, which means that the partial derivatives are continuous. When we move parallel to the axis we are considering a partial derivative, which is what the above graph depicts with dashed lines.
With partial derivatives one variable is treated as a constant, therefore [math]\displaystyle{ G(x) = f(x, y_0 + k) }[/math]. (in case you are confused, the [math]\displaystyle{ x }[/math] to the left is not the same as the one on the right side) With the MVT we can state that there is a [math]\displaystyle{ \bar{x} }[/math], between [math]\displaystyle{ x }[/math] and [math]\displaystyle{ x + h }[/math], such that
[math]\displaystyle{ (\text{I}) = G(x_0 + h) - G(x_0) = G'(\bar{x}) h = \frac{\partial f}{\partial x}(\bar{x}, y_0 + k) h }[/math].
And now we repeat for (II)
[math]\displaystyle{ (\text{II}) = \frac{\partial f}{\partial y}(x_0, \bar{y}) k }[/math].
We began with a point and one increment for the first variable and another increment for the second variable. That is the same concept of a vector, because in analytical geometry we define a vector with two points. If you look back, we have an expression that is similar to the directional derivative, except that we don't care about a specific direction this time around. We care about the continuity of the partial derivatives.
Conceptually, we have reached an equation that represents a vector and the sum of two components that add up to it:
[math]\displaystyle{ f(x_0 + h, y_0 + k) - f(x_0, y_0) = \frac{\partial f}{\partial x}(\bar{x}, y_0 + k)h + \frac{\partial f}{\partial y}(x_0, \bar{y})k }[/math]
Subtract [math]\displaystyle{ \frac{\partial f}{\partial x}(x_0,y_0)h + \frac{\partial f}{\partial y}(x_0,y_0)k }[/math] from both sides of the above equation and so some algebraic operations:
[math]\displaystyle{ f(x_0 + h, y_0 + k) - f(x_0, y_0) - \frac{\partial f}{\partial x}(x_0,y_0)h - \frac{\partial f}{\partial y}(x_0,y_0)k = }[/math]
[math]\displaystyle{ \left[\frac{\partial f}{\partial x}(\bar{x}, y_0 + k) - \frac{\partial f}{\partial x}(x_0,y_0)\right] h + \left[\frac{\partial f}{\partial y}(x_0, \bar{y}) - \frac{\partial f}{\partial y}(x_0,y_0)\right]k }[/math]
If you remember the directional derivative and take a closer look, the left side of the equation has the terms [math]\displaystyle{ -Ah - Bk }[/math], where A and B are the rates of change for each direction. Now if we make [math]\displaystyle{ (h,k) \to (0,0) }[/math] the equation above is reduced to [math]\displaystyle{ f(x_0 + h,y_0 + k) - f(x_0,y_0) = 0 }[/math], which is the same idea when we wish to prove that a function is continuous at a point. We also have to remember that a derivative is a limit, then
[math]\displaystyle{ \left|\frac{f(x_0 + h, y_0 + k) - f(x_0, y_0) - \frac{\partial f}{\partial x}(x_0,y_0)h - \frac{\partial f}{\partial y}(x_0,y_0)k}{||(h,k)||}\right| \leq }[/math]
[math]\displaystyle{ \left|\frac{\partial f}{\partial y}(\bar{x},y_0 + k) - \frac{\partial f}{\partial x}(x_0,y_0)\right|\frac{|h|}{\sqrt{h^2 + k^2}} + \left|\frac{\partial f}{\partial y}(x_0,\bar{y}) - \frac{\partial f}{\partial y}(x_0,y_0)\right|\frac{|k|}{\sqrt{h^2 + k^2}} }[/math]
Functions of the form [math]\displaystyle{ \frac{x^2}{x^2 + y^2} }[/math] are limited, much like sine or cosine. The numerator is never going to be greater than the denominator. In case you are confused, [math]\displaystyle{ \sqrt{h^2 + k^2} }[/math] represents the radius of the open ball from the diagram in the beginning. The partial derivatives are continuous at [math]\displaystyle{ (x_0,y_0) }[/math]. Therefore, the right side of the equation should be zero when [math]\displaystyle{ (h,k) \to (0,0) }[/math].
[math]\displaystyle{ \lim_{(h,k) \ \to \ (0,0)} \frac{f(x_0 + h, y_0 + k) - f(x_0,y_0) - \frac{\partial f}{\partial x}(x_0,y_0)h - \frac{\partial f}{\partial y}(x_0,y_0)k}{||(h,k)||} = 0 }[/math]
With this we have proven that [math]\displaystyle{ f }[/math] is differentiable at [math]\displaystyle{ (x_0,y_0) }[/math].
Note: I've followed Hamilton Guidorizzi's textbook. Other books such as Tom Apostol do the proof for the case of n variables.
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