MHB Relationship Between Total Derivatives and Directional Derivatives .... ....

[Math Amateur]

I am reading Tom M Apostol's book "Mathematical Analysis" (Second Edition) ...I am focused on Chapter 12: Multivariable Differential Calculus ... and in particular on Section 12.4: The Total Derivative ... ...I need help in order to fully understand Theorem 12.3, Section 12.4 ...Theorem 12.3 (including its proof) reads as follows:
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Regarding the above Theorem, I am finding it difficult to understand how, when $$T_c(u) = f'(c;u)$$, that a function can have a finite directional derivative $$f'(c;u)$$ for every $$u$$ but may fail to be continuous at $$c$$ ... whereas! ... for a total derivative ... when a function has a total derivative it is continuous ... YET! ... $$T_c(u) = f'(c;u)$$ ...Can someone please explain what is going on ...
Help will be appreciated ...

Peter
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It may help MHB readers of the above post to have access to Section 12.4 on the total derivative ... so I am providing access to the same ... as follows...
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It may also help MHB readers of the above post to have access to Section 12.2 on the directional derivative ... so I am providing access to the same ... as follows...

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Hope that helps ...

Peter

 

[HOI]

I am reading Tom M Apostol's book "Mathematical Analysis" (Second Edition) ...I am focused on Chapter 12: Multivariable Differential Calculus ... and in particular on Section 12.4: The Total Derivative ... ...I need help in order to fully understand Theorem 12.3, Section 12.4 ...Theorem 12.3 (including its proof) reads as follows:Regarding the above Theorem, I am finding it difficult to understand how, when $$T_c(u) = f'(c;u)$$, that a function can have a finite directional derivative $$f'(c;u)$$ for every $$u$$ but may fail to be continuous at $$c$$ ... whereas! ... for a total derivative ... when a function has a total derivative it is continuous ... YET! ... $$T_c(u) = f'(c;u)$$ ...Can someone please explain what is going on …

Consider the function f(x, y)= 0 if xy= 0, f(x,y)= 1 otherwise. xy= 0 if and only if x= 0 or y= 0. Think of this as the plane z= f(x,y)= 1 with (x, 0, 1) and (0, y, 1), above x and y axes, lowered down to z= 0. At any point on the x-axis the derivative in the x-direction exists (and is 0). At any point on the y-axis the derivative in the y-direction exists and is 0. At the origin, the derivative in both x and y directions exist. But the function is not continuous on the axes.

 

[Opalg]

I am finding it difficult to understand how ... a function can have a finite directional derivative $$f'(c;u)$$ for every $$u$$ but may fail to be continuous at $$c$$.

An example of this is given by the function $f(x,y) = \dfrac{x^2y}{x^4+y^2}$ for all $(x,y)\ne(0,0)$, with $f(0,0) = 0.$ This function has a directional derivative in every direction at the origin. In fact, in the direction of the vector $(u,v)$, \[ \frac{f(hu,hv) - f(0,0)}h = \frac{h^3u^2v}{h(h^4u^4 + h^2v^2)} = \frac{u^2v}{h^2u^4 + v^2}.\] Taking the limit as $h\to0$, \[\lim_{h\to0}\frac{u^2v}{h^2u^4 + v^2} = \begin{cases}0&\text{if }v=0,\\ \frac{u^2}v&\text{if }v\ne0. \end{cases} \] So that limit exists for every direction $(u,v).$

On the other hand, $f(x,x^2) = \dfrac{x^4}{2x^4} \to \frac12$ as $x\to0$. So there are points arbitrarily close to $(0,0)$ at which the function takes the value $\frac12$. Therefore $f$ is not continuous at the origin. Geometrically, what is happening here is that the function tends to $0$ as you approach the origin from any direction in a straight line. But if you approach the origin along the curve $y=x^2$, the function tends to $\frac12$ rather than $0$.

Of course, this function cannot have a total derivative at the origin, because that would imply continuity there.

 

[Math Amateur]

An example of this is given by the function $f(x,y) = \dfrac{x^2y}{x^4+y^2}$ for all $(x,y)\ne(0,0)$, with $f(0,0) = 0.$ This function has a directional derivative in every direction at the origin. In fact, in the direction of the vector $(u,v)$, \[ \frac{f(hu,hv) - f(0,0)}h = \frac{h^3u^2v}{h(h^4u^4 + h^2v^2)} = \frac{u^2v}{h^2u^4 + v^2}.\] Taking the limit as $h\to0$, \[\lim_{h\to0}\frac{u^2v}{h^2u^4 + v^2} = \begin{cases}0&\text{if }v=0,\\ \frac{u^2}v&\text{if }v\ne0. \end{cases} \] So that limit exists for every direction $(u,v).$

On the other hand, $f(x,x^2) = \dfrac{x^4}{2x^4} \to \frac12$ as $x\to0$. So there are points arbitrarily close to $(0,0)$ at which the function takes the value $\frac12$. Therefore $f$ is not continuous at the origin. Geometrically, what is happening here is that the function tends to $0$ as you approach the origin from any direction in a straight line. But if you approach the origin along the curve $y=x^2$, the function tends to $\frac12$ rather than $0$.

Of course, this function cannot have a total derivative at the origin, because that would imply continuity there.

Thanks Country Boy and Opalg

Peter