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Homework Help: So embarrased to ask but i hope one of you math gods will take pity on me

  1. May 5, 2010 #1
    I was wondering if i had a derivative
    deltay/deltax
    where y=x^2+y^2
    would the derivative of the function y be 2x+y^2? or would y also be subject to differentiation?
    sorry guys i know this is simple but any help would be great
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 5, 2010 #2
    sorry i forgot to mention i think the new function y=2x because y^2 is seen as a constant and as such is treated as being 0
     
  4. May 5, 2010 #3
    The answer you supplied is incorrect. I think you should revisit the topic about implicit differentiation. An important point about this is the use of the so called "chain rule". If you use these steps correctly, you will surely get the answer for [itex]y'[/itex] in terms of [itex]x[/itex] and [itex]y[/itex] in 2 lines.
     
  5. May 5, 2010 #4

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    There are two ways to solve problems like this. You can either solve for y, and differentiate (e.g. if you had the function y=2x-2y you can change this to 3y=2x and then to y=2x/3). If you cannot, or it's too difficult, then you must use implicit differentiation as dickfore mentioned. In simple terms you take the derivative w.r.t. x on all 3 terms separately. When you do it to y, you get dy/dt. When you do it to x^2 you get 2x. What happens when you do it to y^2? Hint: It's OK to have a dy/dx term in there!
     
  6. May 5, 2010 #5
    Ditto... the easiest approach to finding dy/dx would be to use implicit differentiation. Review the topic, give it a try and let us know what you get.
     
  7. May 6, 2010 #6
    so would the function be

    dy/dx = 2x+dy^2/d^2x?
     
  8. May 6, 2010 #7

    Mark44

    Staff: Mentor

    No.

    Your derivative for y^2 is incorrect. You haven't used the chain rule correctly.
    [tex]\frac{d}{dx}y^2 = \frac{d}{dy}(y^2) \frac{dy}{dx}[/tex]

    Can you continue from here?
     
  9. May 6, 2010 #8
    Think of it this way: When you differentiate a function, at this level you will be differentiating it with respect to x. dy/dx means(the way I think of it): the derivative of y with respect to x.

    So when you take the derivative of, let's say 2x, you are taking the derivative of x with respect to x, dx/dx, which is just 1. So the derivative of 2x is 2(dx/dx) = 2.

    It also holds for something like x^3. The derivative of x^3 is actually 3x^2 (dx/dx), which is just 3x^2. You usually don't write dx/dx down since it is 1 and should be understood to be that.

    With implicit differentiation, it's the same idea. When you want to take the derivative of something like 1= 2x + 2y implicitly, the derivative will be 0 = 2(dx/dx) + 2(dy/dx) = 2 + 2(dy/dx). dy/dx should look familiar since it's the notation for the derivative, so you would solve for dy/dx. Hopefully this makes sense and should help you solve the problem.
     
  10. May 6, 2010 #9

    Mark44

    Staff: Mentor

    And it is often read this way.
    You are taking the derivative of 2x with respect to x; i.e., d(2x)/dx. If you want to look at numerator and denominator separately, using differentials, you have d(2x)/dx = 2dx/dx = 2. This seems a needless complication, IMO.
    I would imagine that the OP is already familiar with the fact that d(x^3)/dx = 3x^2. Adding the factor of dx/dx is again a needless complication, IMO.
     
  11. May 6, 2010 #10
    Sorry, I was just trying to show why you need a dy/dx when implicitly differentiating after differentiating a term with y. It makes it a little more complicated, but I thought it would make more sense to help see the problem in that way.
     
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