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Proof not making sence to me.

  • Thread starter Miike012
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  • #1
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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 thats obviously not what it means... because they have y = cu? so can some one explain??????
 
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Answers and Replies

  • #2
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).
 
  • #3
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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?
 
  • #4
gb7nash
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would it be y = cx ; and u = cx? But thats 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?
 
  • #5
Char. Limit
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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).
 
  • #6
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.
 
  • #7
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Yes.. c Is a constant.
I just dont 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 doesnt y' = c ?
 
  • #8
gb7nash
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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.
 
  • #9
Char. Limit
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Yes.. c Is a constant.
I just dont 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 doesnt 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).
 
  • #10
gb7nash
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  • #11
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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 dont know what g(x) is in the first place?
 
  • #12
Char. Limit
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[tex]g(x+h)-g(x) = \Delta g[/tex], and [tex]x+h-x = \Delta x = h[/tex]
 
  • #13
gb7nash
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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?
 
  • #14
Char. Limit
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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 dont 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)
 
  • #15
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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.
 
  • #16
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My way I can find the logic but I cant see the logic in the other way.
 
  • #17
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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
 
  • #18
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csdsdsdsd
 
  • #19
Char. Limit
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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...

[tex]lim_{h\rightarrow0} \frac{g(x+h)-g(x)}{h} = g'(x)[/tex]

All you need for the proof is this and how to factor out a c.
 
  • #20
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factor out a "c" ? I dont see a c to factor out?
 
  • #21
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Nevermind... im guessing your referring to my original prob.?
 
  • #22
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Yes... I know the equation for finding a derivative... but im pretending that I dont know it... im trying to go about it a diff way... useing the equation that you presented can be done by anyone.
 
  • #23
Char. Limit
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Remember that your original statement was...

[tex]y(x) = c u(x)[/tex]

and you wanted to prove that

[tex]y'(x) = c u'(x)[/tex]

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

[tex]y'(x) = lim_{h\rightarrow0} \frac{y(x+h) - y(x)}{h} = lim_{h\rightarrow0} \frac{c u(x+h) - c u(x)}{h}[/tex]

Any more help than that and I'd probably be violating the rules.
 
  • #24
HallsofIvy
Science Advisor
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Yes.. c Is a constant.
I just dont 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 doesnt 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.
 
  • #25
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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' ?
 
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