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Chain Rule

  • Thread starter Miike012
  • Start date
  • #1
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Homework Statement


If x1 is some value of x, u1is the corresponding value of u and y1 the corresponding value of y, then dy/dx at x = x1 is the product of dy/du at u1 and du/dx at x1.

This does not make sence....
example
IF: y = (x+2)^2
y=u^2 u = x+2

dy/du = 2u
du/dx = 1

I want to find the tangent at x = 2

The above statment says dy/dx = dy/du at point u1 time du/dx at point x1

dy/du @ x = 2 is 4

du/dx @ u =2 is 1

4*1 is 4 which is not the correct answer.....



The Attempt at a Solution

 

Answers and Replies

  • #2
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are you trying to find the equation of the line tangent to y=(x+2)^2 at the point x=2?
 
  • #3
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Yes. I wanted to know how to do it using the chain rule... but I figured it out...
However I ran into another problem, maybe you can help.

It says.... find the derivative of y^2 with respect to x using the chain rule.

The chain rule is...dy/dx = dy/du * du/dx

If I substitue values in I get.... dy^2/dx = dy/du * du/dx ( what do I substitue in for dy/du * du/dx ?)
 
  • #4
196
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That sounds like implicit differentiation, if it is then if you want to differentiate y with respect to x you let y be defined implicitly by x which would look something like this:

(f(x))^2, where f(x)=y

So if you use the chain rule you would have 2(f(x))*f'(x) which you can rewrite in terms of since f(x)=y
 
  • #5
1,011
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In the book it says... differnentiate y^2 then multiply is by dy/dx
= 2y dy/dx

Why do we multiply it by dy/dx?
 
  • #6
19
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so tangent at x=2 for y=(x+2)^2 ... if U=x+2 then you had it partially right, you got the 2U which is the 2(x+2) ... but because you are substituting U in you have to treat U as a function which means you would have 2U(U')
 
  • #7
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In x=y^2, x is a function of y. So if you want to differentiate y with respect to x, you have to redefine y in terms of x: y=(f(x))^2. This is where it becomes a chain rule problem since you have a function within a function: x^2 is the outer function and f(x) is the inner function. So you multiply y^2 by dy/dx which is the same thing as f'(x)

So in your case, the derivative of y^2 with respect to x is 2yy' or 2(f(x))*f'(x).
 
  • #8
1,011
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Thank you.. I understand it now.
 
  • #9
1,011
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so the original was x = y^2
then it looks like you changed y^2 to x^2 and because you said "y" is the inside I am assuming that I could set y = x
so now that I made the change is the new equation y = x^2 ?

Then dy^2/dx = dy/dx * dx/dy
(I dont think I am correct?)
anyways... dx/dy = 2x an because y=x with substitution it becomes 2y...

now we have dy^2/dx = 2y* dy/dx

Is this correct? I think I am doing something wrong because if I set y = x that would mean dy/dx is 1
then that would give us dy^2/dx = 2y* 1
 
  • #10
verty
Homework Helper
2,164
198
Miike012 said:
so the original was x = y^2
then it looks like you changed y^2 to x^2 and because you said "y" is the inside I am assuming that I could set y = x
so now that I made the change is the new equation y = x^2 ?
Let's see. x = y^2. If you want dy/dx, you need to solve for y: y = x^(1/2) or y = -x^(1/2). Then dy/dx = .....
 
  • #11
1,011
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But its not east to solve for x
what if it was y^3 - xy^2 +2x - y... would you use the chain rule?
 
  • #12
196
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when you differentiate y with respect to x you are redefining y as some function of x, say f(x). And since it is y^2 it would become (f(x))^2 where the chain rule now applies since you have a function within a function.

And yes, if it was Y^3-xy^2+2x-y=0 you would still use the chain rule on each y term:

(f(x))^3-x(f(x))+2x-f(x)

every time there is an f(x) you can apply the chain rule
 

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