What is the Relationship Between Differentiability and Limits?

[Miike012]

If y and u are functions of x and y = cu
then y' = cu'

What does "If y and u are functions of x " look like on paper?

would it be y = cx ; and u = cx? But that's obviously not what it means... because they have y = cu? so can some one explain?

 

[╔(σ_σ)╝]

Typically when we say y is a function of x it simply means y =f(x) for some map f.
i.e

y= sinx
y= ax
y = cos^-1(sin(tan(x)) etc

Y need not be a constant multiple of x.

So in your case you simply have y= f(x) and u =g(x).

 

[Miike012]

I am assuming that the original equation was y = cx
Then they stated Let y and u be functions of x.
That is how we got y = cu, correct?
Anyways... How did they get from y = cx to y = cu. Did they use the information that " y= f(x) and u =g(x)" and manipulate it somehow?

 

[gb7nash]

would it be y = cx ; and u = cx? But that's obviously not what it means... because they have y = cu? so can some one explain?

Is this part of the problem? I'm not sure where you're getting y = cx from. Also, is c a constant?

 

[Char. Limit]

What they are saying is that if u=f(x), and y=c*f(x) for some constant c, then y'=c*u'=c*f'(x).

 

[╔(σ_σ)╝]

I am assuming that the original equation was y = cx
Then they stated Let y and u be functions of x.
That is how we got y = cu, correct?
Anyways... How did they get from y = cx to y = cu. Did they use the information that " y= f(x) and u =g(x)" and manipulate it somehow?

The way I understand it is as follows...
y= f(x) and u=g(x)

y= cu is the same as f(x) = c*g(x).

Basically, I understand this to mean that f(x)( or y) is simpley g(x) (or u) multiplied by c ( is c a constant ? I assume so.)

For example if

u = sinx then
y = c*sinx

y' = c*cosx.

 

[Miike012]

Yes.. c Is a constant.
I just don't understand the proof that...
if y = cx... then y' equals a constant c times a derivative...

The theorem says that... y = cx... then y' = cx'
But why doesn't y' = c ?

 

[gb7nash]

If that's the case, I'm not sure what there is to prove here. y' = cf'(x) follows straight from a property of differentiation, but I'm not sure how far you are in derivatives.

 

[Char. Limit]

Yes.. c Is a constant.
I just don't understand the proof that...
if y = cx... then y' equals a constant c times a derivative...

The theorem says that... y = cx... then y' = cx'
But why doesn't y' = c ?

No, it doesn't. The theorem does not define y as y=cx. It defines y as y=cu, where u=u(x).

 

[gb7nash]

https://www.physicsforums.com/showthread.php?t=258405

Taken from the other thread, let f(x) = cg(x). So using the limit definition:

[lim h->0] (f(x+h)-f(x))/h = [lim h->0] (cg(x+h)-cg(x))/h = ...

 

[Miike012]

What does


g(x + h) - g(x) =?

I understand if say...
g(x) = x^2

then
g(x + h) - g(x) = (x+h)^2 - x^2

So what would g(x + h) - g(x) equal if I don't know what g(x) is in the first place?

 

[Char. Limit]

$$g(x+h)-g(x) = \Delta g$$, and $$x+h-x = \Delta x = h$$

 

[gb7nash]

What does


g(x + h) - g(x) =?

What is the definition of g'(x)? How can we get that from the last thing I wrote?

 

[Char. Limit]

What does


g(x + h) - g(x) =?

I understand if say...
g(x) = x^2

then
g(x + h) - g(x) = (x+h)^2 - x^2

So what would g(x + h) - g(x) equal if I don't know what g(x) is in the first place?

Well, g(x+h)-g(x) isn't equal to anything by itself. However, you can say that lim h-->0 g(x+h)-g(x) = h g'(x)

 

[Miike012]

This is the way I am used to proving it...

Say: Y = x^2
We compute the first value of y when x has the value of x1. This value of y which will be called y1 is obtained by sub x1 for x..
Hence y1 = (x1)^2

Then we sub x1 + delta(x) with the corresponding y value y1 + delta(y)
y1 + delta(y) = ( x1 + delta(x))^2

Now subtract them
delta(y) = 2x1 * Delta(x) + (delta(x))^2
Divide by delta(x) then find lim at delta(x) --> 0
dy/dx = 2x.

 

[Miike012]

My way I can find the logic but I can't see the logic in the other way.

 

[Miike012]

g(x+h)-g(x) = h g'(x)


The only thing that I can think of that you are doing is
adding x + h - x = h
then divide by h = 1
then find the limit as g ---> 0 which equals g'.
What the heck? that makes no sence

 

[Miike012]

csdsdsdsd

 

[Char. Limit]

g(x+h)-g(x) = h g'(x)


The only thing that I can think of that you are doing is
adding x + h - x = h
then divide by h = 1
then find the limit as g ---> 0 which equals g'.
What the heck? that makes no sence

Well, basically I'm using the fact that...

$$lim_{h\rightarrow0} \frac{g(x+h)-g(x)}{h} = g'(x)$$

All you need for the proof is this and how to factor out a c.

 

[Miike012]

factor out a "c" ? I don't see a c to factor out?

 

[Miike012]

Nevermind... I am guessing your referring to my original prob.?

 

[Miike012]

Yes... I know the equation for finding a derivative... but I am pretending that I don't know it... I am trying to go about it a diff way... useing the equation that you presented can be done by anyone.

 

[Char. Limit]

Remember that your original statement was...

$$y(x) = c u(x)$$

and you wanted to prove that

$$y'(x) = c u'(x)$$

Well, you just use the definition of the derivative for y(x).

$$y'(x) = lim_{h\rightarrow0} \frac{y(x+h) - y(x)}{h} = lim_{h\rightarrow0} \frac{c u(x+h) - c u(x)}{h}$$

Any more help than that and I'd probably be violating the rules.

 

[HallsofIvy]

Yes.. c Is a constant.
I just don't understand the proof that...
if y = cx... then y' equals a constant c times a derivative...

The theorem says that... y = cx... then y' = cx'
But why doesn't y' = c ?

The prime indicates the derivative with respect to what variable?

If you are differentiating with respect to x itself, then, yes, y'= (cx)'= c. But if x is a function of some other variable, and you are differentiating with respect to that other variable, then, by the chain rule, y'= cx'. In fact, if you are differentiating with respect to x itself, x'= 1 so the equation y'= cx' is true no matter what variable you are differentiating with respect to.

 

[Miike012]

Sorry HallsofIvy.. I got lost... you said
If you are differentiating with respect to x itself, then, yes, y'= (cx)'= c
Then you said
x'= 1 so the equation y'= cx'

So you are saying that y' = c is equal to y' = cx' ?

 

[Miike012]

This proof is almost as if the person made this proof was like...
If there is a function say x^3 + c OR x^3
Then by differentiating , both functons would be 3x^2
Now the question pops up... If we had a differentiated function f'(x) without knowing the original function f(x)...
How could we know that the integration of f'(x) differs by only a constant...
Well if we set
x^3 + c = x^3
then by diff..
= 3x^2 + c' = 3x^2
which equals
c' = 0...
Am I correct?

 

[jgens]

However, you can say that lim h-->0 g(x+h)-g(x) = h g'(x)

No. If g is differentiable at x, then you have that

$$\lim_{h \to 0}[g(x+h)-g(x)] = \lim_{h \to 0} hg'(x)$$

 

[Miike012]

...

 

[Char. Limit]

No. If g is differentiable at x, then you have that

$$\lim_{h \to 0}[g(x+h)-g(x)] = \lim_{h \to 0} hg'(x)$$

Sorry, I intended the limit to extend to both sides of the inequality.