1. Jun 23, 2015

### christian0710

Hi, my Calculus book (Elementary Calculus an Infinitesimal approach) uses a mathematical argument, and I'm not sure if it's correct so i thought i wuold check with you guys :)

When the book want's to find the derivative of f(x)=x^2 the books starts like this

y=x2
Δy +y = (x + Δx)2

So my question is this: Is it mathematically correct to write
Δy+y = (x + Δx)2.

To me it looks more like the left side should translate into a right side that looks like this
Δy+y = x2 + (Δx)2

I'm asuming that Δy+y translates into f(Δx+x) and not f(x)+f(Δx) but to me it looks as if Δy+y is f(x)+f(Δx) which is not the same as f(Δx+x)

If the above is true, does that mean that this is also true?

if y=x2 then.
y=x2
y+1/y = (x+1/x)2 and would this be the same as f(x+1/x)?

Last edited by a moderator: Jun 23, 2015
2. Jun 23, 2015

### fireflies

No, here x2 is a function of x
That is y=f(x)=x2

3. Jun 23, 2015

### Staff: Mentor

Yes.
No, not at all. $(x + \Delta x)^2 = x^2 + 2x\Delta x + (\Delta x)^2$, not x2 + (Δx)2 as you have.
Δy+y = f(Δx+x) ≠ f(Δx) + f(x).
You are mistakenly thinking that the distributive property applies here, which it does not.
The distributive property applies when you multiply a sum of numbers, as in a(b + c) = ab + ac. It does NOT apply to functions, in general. For example, $\sqrt{a + b} \neq \sqrt{a} + \sqrt{b}$, and $\sin(x + y) \neq \sin(x) + \sin(y)$.
Let's define y = f(x) = x2

What is y + 1/y?
What is f(x + 1/x)?
Are they equal?

4. Jun 23, 2015

### fireflies

Del means something like a little change of..

So here a little change of x is added. For example, when you put x=2 then y = 4,
With a little change in x like del x=1 then x will be equal to 5, and y will be equal to 52=25, not (4)2+(1)2

Last edited: Jun 23, 2015
5. Jun 23, 2015

### christian0710

if y = f(x) = x2
Then y+1/y = (x+1/x)^2 which is the same as f(x+1/x) Correct?

So if i take f(f(x+1/x)) is it then the same as f(x+1/x) = (x+1/x)^2 ? or is it ( (x+1/x)+1/(x+1/x))2?

6. Jun 23, 2015

### Staff: Mentor

No. Let's start from the other end.
f(x + 1/x) = (x + 1/x)2, which is what you have.
But (x + 1/x)2 = x2 + 2 * x * 1/x + 1/x2 = x2 + 2 + 1/x2 = f(x) + 2 + f(1/x)

The last expression is not equal to either y + 1/y or f(x) + f(1/x).

7. Jun 24, 2015

### christian0710

Then I think I'm misunderstanding one very important thing, so let's see where my logic is wrong:

Above we agreed that
Δy+y = f(Δx+x), So i assume this means "So if you add some y or Δy or 1/y to the left side it's the same as putting it into the f(x) operator, so if the left side says
Δy+Δy+y then this is equal to f(Δx+Δx+y) and therefor it should also be true that y+1/y = f(x+1/x) So I'm erroneously treating Δy like y?

Just to make sure, could this be one of my wrong ways of thinking ?
If f(x)=y=1/x then
y+Δy = 1((x+Δx) which is true BUT now I'm erroneously thinking that if you put something on the left side, it goes into the "x" on the right side, and therefore:
y+y = 1/(x+x) (I think I'm treating γ like Δy) while in fact it should be y+y = 1/x+1/x =2/x

So my question is, why don't we treat Δy and y the same way? Why is it not Δy+y = 1/x +1/Δx

UPDATE: Is this because Δy means "An increment in y" while y means "A complete function", so by adding 2*Δy on the left side we are not adding 2 functions on the right, but we are incrementing x to 2*Δx and this is the same as (x + 2*Δx) :D

Last edited: Jun 24, 2015
8. Jun 24, 2015

### Staff: Mentor

In general, no.

In your first post, you started with f(x) = x2, so lets stick with that.
You should sketch a graph of this function, with two points labeled: A(x, x2) and B(x + Δx, (x + Δx)2). The horizontal distance between A and B is Δx. The vertical distance between A and B is Δy = (x + Δx)2 - x2 = 2xΔx + (Δx)2. If we add a third point C(x + 2Δx, (x + 2Δx)2), the vertical distance between B and C will NOT be Δy.

BTW, f is a function, not an operator. f(x) is the number that the function associates with a value x in the domain of f.
No, it's not true, due to a typo.

Since you started off this thread with f defined as f(x) = x2, you should use a different letter if you now want to work with a different function.

Let's say g(x) = 1/x
So g(x + Δx) = $\frac{1}{x + \Delta x}$
$\Delta y = g(x + \Delta x) - g(x) = \frac{1}{x + \Delta x} - \frac{1}{x}$
Δx is a symbol that represents a "small" change in x. Δy similarly represents a small change in y. More precisely, Δy = f(x + Δx) - f(x).

The main problem, as I see it, is that you are focussing on the left side without understanding what the right side of your equation means.

9. Jun 24, 2015

### PeroK

You're getting yourself in a real old tangle here! First (by definition):

$\Delta y = y(x + \Delta x) - y(x)$

Whereas, I think you want to define:

$\Delta y = y(\Delta x)$

Last edited: Jun 24, 2015
10. Jun 24, 2015

### fireflies

It's something like y=f(x), so, /frac{1}{y}=/frac {1}{f(x)}

In a function you should put the value of the variable of which the function is of. In case of f(x) for new values of x (which may be x+del x, or 1 or 3 or any such) you should put the value in the function.

11. Jun 24, 2015

### christian0710

+
YES I think I finally understand it now, I just couldn't see the full argument : Do we agree on this:

The book says f(x) = x2

y = y(x) =x^2

and when we add Δy, we are actually doing this

y +Δy = y(x) +(y(x+Δx) -y(x)) = y(x+Δx) = (x+Δx)2

I couldn't understand that when we add Δy on the left side, it looked like it suddently appeared inside y(x) as y(x+Δx). beucase all i saw on the left side was, that when you add Δy on the left side, it appears inside x as (x+Δx) so i thought that because Δy+y = f(x+Δx) then it must also be true that y+y = f(x+x) but Mark44 helped me se that it's not the case and the reason that y +Δy = y(x+Δx) is because y +Δy = y(x) +(y(x+Δx) -y(x)) = y(x+Δx)

Thank you guys :D

12. Jun 26, 2015

### HallsofIvy

Staff Emeritus
The point is that they are trying to answer "if we increase x a small amount, what happens to y?" If $y= x^2$ and we increase x by the small amount $\Delta x$, then y is changed to $(x+ \Delta x)^2= x^2+ 2(\Delta x)x+ (\Delta x)^2$. Since that $x^2$ is just y itself, this is the same as $y+ 2(\Delta x)x+ (\Delta x)^2$. So increasing x by $\Delta x$ increases y by $2(\Delta x)x+ (\Delta x)^2$ which, since it is a small increase in y, we call $\Delta y$.