Solving y' in 2x+2xy'+2y+3y2y'

  • Thread starter Centurion1
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In summary, the homework statement has two equations and one unknown. The Attempt at a Solution states that y' is treated like another variable and can be factored. Homework Equations state that two equations need to be solved for two unknowns.
  • #1
Centurion1
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Homework Statement


Solve for y'

2x+2xy'+2y+3y2y'


Homework Equations





The Attempt at a Solution



so the ' means degrees right?
Would you start by grouping like terms. My real question is how that ' makes the other variables change?
 
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  • #2
No, [tex] y' [/tex] is the first derivative of [tex] y [/tex]. You need to isolate [tex] y' [/tex] from the other variables, but I see one problem: you don't have this expression set equal to anything: as it stands you can factor it (a little), but there is no way to solve it for any of the quantities.
 
  • #3
You need to be more specific. To solve for a variable you need to have an equation. What you have written down is not an equation at all so make sure you also write down the other side of the equation like so, 'what you have'=...

Usually y' means the derivative of y. In this case I would guess with respect to x.
 
  • #4
sorry it is =0
 
  • #5
Centurion1 said:
sorry it is =0

Then, factor out y' and solve the equation.
 
  • #6
You mean, 2x+2xy'+2y+3y2y' = 0, and want to find y'.

You can precisely perform steps according to:

Commutative property of addition;
Additive Inverse property;
Like Terms or Reverse of Distributive Property;
Multiplicative Inverse property.
 
  • #7
so y' is treated like another variable. I can do normal factoring and just treat it like it was a third variable like if it was a z instead?
 
  • #8
Centurion1 said:
so y' is treated like another variable. I can do normal factoring and just treat it like it was a third variable like if it was a z instead?

YES, from the viewpoint of "Pre-Calculus" and any Math course below. Some variable names may have subscripts to distinguish between or among similar main symbols. The "prime" notation can also be used sometimes. You have the variable written, y', spoken as "y prime". Your example also has the variable, y. So, two of your variables in your example are y and y' ("y" and "y prime"). We assume that they share some character but that they represent different numbers.
 
  • #9
yes i understand now, thank you. i think i can handle the factoring.
 

FAQ: Solving y' in 2x+2xy'+2y+3y2y'

What does "Solving y' in 2x+2xy'+2y+3y2y'" mean?

Solving y' in this equation means finding the expression for the derivative of y with respect to x, which is represented by y'.

How do you solve for y' in 2x+2xy'+2y+3y2y'?

To solve for y', you must use the rules of differentiation to find the derivative of each term in the equation. Once you have found the derivative, set it equal to y' and solve for y'.

Why is it important to solve for y' in this equation?

Solving for y' allows us to find the slope of the tangent line to the curve represented by the equation at any given point. This is useful in many applications, such as finding maximum and minimum points, and determining the rate of change of a function.

Can you provide an example of solving y' in 2x+2xy'+2y+3y2y'?

Sure, let's say we have the equation y = x^2 + 3x. To solve for y', we first find the derivative of each term: y' = 2x + 3. Then, we set y' equal to the given expression, 2x+2xy'+2y+3y2y': 2x+2xy'+2y+3y2y' = 2x + 3. Solving for y', we get y' = (3-2xy')/(2+3y).

Are there any special cases to consider when solving for y' in this equation?

Yes, there are a few special cases to consider. If the equation contains trigonometric functions, logarithmic functions, or exponential functions, you will need to use the appropriate derivative rules for those functions. Additionally, if the equation contains multiple variables, such as x and y, you will need to use the chain rule to find the derivative of y with respect to x.

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