# Directional Derivative

1. Aug 1, 2005

### Cyrus

Stewart, page 800 says:

"Proof: If we define a function g of the single variable h by

$$g(h) = f(x_0 + ha, y_0 + hb)$$

then by the definition of a derivative we have

$$g'(0)= lim_{h \rightarrow 0} \frac{g(h) - g(0)}{h} = lim_{h \rightarrow 0} \frac{f(x_0+ha, y_0+hb)-f(x_0,y_0)}{h}$$

end quote

Is it me, or is he over using the variable h? He defines a function called g(h). And then he puts h back into the derivative. if h=0, then it does not make sense to say g(h)-g(0), becuase he said before that h=0. Should he call the function g(h), g(h'), and then he can call the h in the limit, plain old h?

2. Aug 1, 2005

### hypermorphism

If you changed h to h' in the numerator, what would be the use of studying the behavior of the quotient as h (a variable that now only appears in the denominator) approached zero ?

3. Aug 1, 2005

### Cyrus

No, im saying he should write it as this :

$$g(h') = f(x_0 + h'a, y_0 + h'b)$$

In the single variable case, he said

$$f'(x) = lim_{ h \rightarrow 0} \frac{ f(x+h) - f(x)}{h}$$

Here he called the f'(x) part a variable x, not h. Yet he seems to call it g(h), and then re-use the variable h inside his limit.

4. Aug 1, 2005

### Cyrus

It makes more sense to me for him to say,

$$g(\zeta)=f(x_0+\zeta a,y_0 + \zeta b)$$

Where zeta is some arbitrary variable that bears no resemblance to x or y or z.

Then he could say

$$g'(\zeta) = lim_{h \rightarrow 0} \frac{g(\zeta + h) - g(\zeta)}{h}$$

Then all you have to do is plug in zero for zeta, giving you:

$$g'(0) = lim_{h \rightarrow 0} \frac{g(0+ h) - g(0)}{h} = lim_{h \rightarrow 0} \frac{g(h) - g(0)}{h}$$

Which is what stewart has, quite sloppily.

And then that is equal to:

$$lim_{h \rightarrow 0} \frac{g(h) - g(0)}{h} = lim_{h \rightarrow 0} \frac{f(x_0+(\zeta + h)a, y_0 + (\zeta+h)b - f(x_0+\zeta a,y_0+ \zeta b)} {h}$$

again, plugging in zero gives us what we want:

$$lim_{h \rightarrow 0} \frac{g(h) - g(0)}{h} = lim_{h \rightarrow 0} \frac{f(x_0+ha, y_0 + hb) - f(x_0,y_0)} {h}$$

Now im almost convicned he over used that variable h quite loosely

Last edited: Aug 1, 2005
5. Aug 2, 2005

### Galileo

I agree he shouldn't use the variable h for different purposes. It could lead to confusion if you don't know what he's talking about. If you DO know it won't be a problem, so it doesn't matter anyway. What Stewart wrote isn't wrong.

$$g'(0)= \lim_{h \to 0}\frac{g(h)-g(0)}{h}= \lim_{h \to 0} \frac{f(x_0+ha,y_0+hb)-f(x_0,y_0)}{h}=D_{\vec u} f(x_0,y_0)$$
is a number that is independent of h. Doesn't matter what you use in place of it.
If you use h, then you can place the expression for g(h) in the numerator immediately, but I agree he should've used something else.

6. Aug 2, 2005

### Hurkyl

Staff Emeritus
Nope. (At least, not in the sense of writing correct mathematics)

No, he defines a function called g.

This is incorrect. In that post, you defined the function g via $g(\zeta)=f(x_0+\zeta a,y_0 + \zeta b)$. Thus, we have, for example, that $g(h)=f(x_0+h a,y_0 + h b)$

7. Aug 2, 2005

### Cyrus

What is wrong with that notation? See, the way I used h, it is represents a small change in increment sorta like delta x, its not a matter of setting zeta equal to h. Stewart sometimes uses as h goes to zero in his definition of derivative, and sometimes he uses as x2->x1. I think the problem was that in this case, he was not using h for that purpose. Its sorta confusing because he has two definitions of derivative:

$$f'(x) = lim_{h \rightarrow 0} \frac {f(x+h) - f(x) }{h}$$

but he was using this definition of the derivative:

$$f'(a) = lim_{x \rightarrow a} \frac{f(x+a) - f(a)}{x-a}$$

And replacing the x with the variable h, and the a with the value zero.

So you can see why im saying he kind of used the letter h loosely. Also, he did not write h-0 in the denominator, which lead me to think he was trying to use the first definition of derivative. In which case he would have had the h as both a variable and an incremental value. It would have looked like this

$$f'(h) = lim_{h \rightarrow 0} \frac{f(h+h) - f(h)}{h}$$

what he REALLY meant to write was:

$$f'(0) = lim_{h \rightarrow 0} \frac{f(0+h) -f(0)}{h-0}$$

This makes it clear which definition he is using, becuase now h is not being over used.

Last edited: Aug 2, 2005
8. Aug 2, 2005

### shmoe

You define g by

$$g(\zeta)=f(x_0+\zeta a,y_0 + \zeta b)$$

yet you somehow have:

$$g(0)=f(x_0+\zeta a,y_0+ \zeta b)$$

You don't see a problem with this?

Which are both equivalent of course and look even more similar when a=0. Are you faulting him for not writing "0" in a couple of places so it looks exactly like the derivative formula you want it to look like? Don't expect this kind of hand holding from a maths text, the two definitions are equivalent get used to shifted from one to the other, and get used to authors removing 0's without warning. Everything Stewart wrote is mathematically correct.

I wonder what you would do if someone tried to define x as a function of f? $$x(f)=f^2+\sin(f)$$

9. Aug 2, 2005

### Cyrus

Ah, your right, I did not notice that I had those zetas in there, they should be zeros. I did that in the next line. My fault. It should read something more like this:

$$g'(\zeta) = lim_{h \rightarrow 0} \frac{g(\zeta + h) - g(\zeta )}{h} = lim_{h \rightarrow 0} \frac{f(x_0+(\zeta + h)a, y_0 + (\zeta+h)b - f(x_0+\zeta a,y_0+ \zeta b)} {h}$$

But you do see, that by him using h and not saying h-0, it can make one think he means h in the other sense? It is confusing. Its correct, but leaves some ambiguity on how one interperts the use of h.

Last edited: Aug 2, 2005
10. Aug 2, 2005

### shmoe

If you're reading it just looking at the symbols without really understanding their meaning, then you might be confused on your first pass (not understanding the meaning on your first pass isn't a horrid thing). But when you stop to think about what he's saying, it should be clear what's what.

There is no ambiguity.

11. Aug 2, 2005

### Cyrus

There is if you think hes using the definition based upon:

$$f'(x) = lim_{h \rightarrow 0} \frac{ f(x+h)-f(x)}{h}$$

Becuase he wrote g(h), so the f'(x), would be g'(h), and you also have a DIFFERENT h inside the f(x+h), because you have g(h+h).

Is that not ambiguous?

12. Aug 2, 2005

### shmoe

He is though. I've said these definitions are equivalent and I'm sure you know this. He could have used

$$f'(x) = \lim_{\tau \rightarrow 0} \frac{ (f(x+\tau))^{\cos(0)}-f(\frac{2}{2}x)}{\tau+0^{20}}$$

for his derivative if he wanted.

No it's not. He writes a limit for g'(0). See the zero? There's no h there at all, so having an h in your limit is fine. If he was trying to write a limit expressing g'(h) (<-note the general h here) and h was going to zero in the limit, there would be a problem. But this is distinctly not what he's doing.

13. Aug 2, 2005

### Cyrus

I think we are saying the same thing now. Because as I said before, if he wrote g'(h) there would be a problem. But if he uses g'(a), then its ok. Anyways, I think the problem has been cleared up, thanks :-)

14. Aug 2, 2005

### Hurkyl

Staff Emeritus
The following statement is, in fact, a bad statement:

$$g'(h) = \lim_{h \rightarrow 0} \frac{g(h + h) - g(h)}{h}$$