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Implicit Differentiation

  1. Feb 19, 2017 #1
    1. The problem statement, all variables and given/known data
    [itex]\dfrac{x^2}{x+y}=y^2+8[/itex]

    2. Relevant equations

    Quotient Rule: [itex]\dfrac{g(x)\cdot f'(x)-g'(x)\cdot f(x)}{(g(x))^2}[/itex]

    Product Rule: [itex]f(x)\cdot g'(x)+g(x)\cdot f'(x)[/itex]

    3. The attempt at a solution

    [itex]\dfrac{(x+y\cdot\dfrac{dy}{dx})(2x)-(1\cdot\dfrac{dy}{dx})(x^2)}{(x+y\cdot \dfrac{dy}{dx})^2} = 2y\cdot\dfrac{dy}{dx}[/itex]

    I feel like there are a couple of ways to go about this. Would it be easier to flip the denominator and use the product rule? I just used the quotient from here on out.

    What you see my trying to do is get [itex]\dfrac{dy}{dx}[/itex] on the left side and everything else on the right, then factor [itex]\dfrac{dy}{dx}[/itex] out. This is my first encounter with this type of problem so I get confused very fast.

    [itex]\dfrac{(x+y\cdot\dfrac{dy}{dx})(2x)-(1\cdot\dfrac{dy}{dx})(x^2)}{(x+y\cdot \dfrac{dy}{dx})^2} = 2y\cdot\dfrac{dy}{dx}[/itex]

    [itex]\dfrac{(2x^2+2xy\cdot\dfrac{dy}{dx})-(x^2\cdot\dfrac{dy}{dx})}{(x+y\cdot \dfrac{dy}{dx})^2} = 2y\cdot\dfrac{dy}{dx}[/itex]

    [itex]\dfrac{\dfrac{dy}{dx}-(x^2\cdot\dfrac{dy}{dx})}{(x+y\cdot\dfrac{dy}{dx})^2}=\dfrac{2y\cdot\dfrac{dy}{dx}}{(2x^2+2xy\cdot\dfrac{dy}{dx})}[/itex]

    [itex]\dfrac{-1}{x^2}\cdot\dfrac{\dfrac{dy}{dx}-(x^2\cdot\dfrac{dy}{dx})}{(x+y\cdot\dfrac{dy}{dx})^2}=\dfrac{2y\cdot\dfrac{dy}{dx}}{(2x^2+2xy)}\cdot\dfrac{-1}{x^2}[/itex]

    [itex]\dfrac{\dfrac{dy}{dx}-\dfrac{dy}{dx}}{(x+y\cdot\dfrac{dy}{dx})^2}=\dfrac{2y\cdot\dfrac{dy}{dx}}{-x^2(2x^2+2xy)}[/itex]

    [itex]\dfrac{1}{\dfrac{dy}{dx}}\cdot\dfrac{\dfrac{dy}{dx}-\dfrac{dy}{dx}}{(x+y\cdot\dfrac{dy}{dx})^2}=\dfrac{2y\cdot\dfrac{dy}{dx}}{-x^2(2x^2+2xy)}\cdot\dfrac{1}{\dfrac{dy}{dx}}[/itex]

    [itex]\dfrac{\dfrac{dy}{dx}-\dfrac{dy}{dx}}{\dfrac{dy}{dx}\cdot(x+y\cdot\dfrac{dy}{dx})^2}=\dfrac{2y}{-x^2(2x^2+2xy)}[/itex]

    I don't want to go much farther because I could be doing this wrong. On the left side, I want to factor, but I'm curious if this is right so far or have I made any errors?
     
  2. jcsd
  3. Feb 19, 2017 #2

    Ray Vickson

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    Where does the ##y y'## come from on the left?
     
  4. Feb 19, 2017 #3
    Are you talking about where I have [itex](x+y\cdot \dfrac{dy}{dx})[/itex] or [itex]-(1\cdot \dfrac{dy}{dx})[/itex]?
     
  5. Feb 19, 2017 #4

    ehild

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    The differentiation of the left side is wrong. What are f and g ? Just apply the Quotient Rule properly.
     
  6. Feb 19, 2017 #5
    Oops, here they are:
    [itex]f(x)=x^2[/itex]
    [itex]f'(x)=2x[/itex]
    [itex]g(x)=(x+y)[/itex]
    [itex]g'(x)=(x+y\cdot\dfrac{dy}{dx})[/itex]

    Edit: Oh, I need it to say [itex]\dfrac{(x+y)(2x)-(1\cdot\dfrac{dy}{dx})(x^2)}{(x+y\cdot \dfrac{dy}{dx})^2} = 2y\cdot\dfrac{dy}{dx}[/itex]
     
  7. Feb 19, 2017 #6

    ehild

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    The last equation is wrong. The derivative of a sum is the sum of derivatives. What is dx/dx? and dy/dx is not yy'.
     
  8. Feb 19, 2017 #7
    Would it just be [itex]\dfrac{dy}{dx}[/itex]
     
  9. Feb 19, 2017 #8

    ehild

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    Still wrong. What should be the denominator? Is not it g2?
     
  10. Feb 19, 2017 #9

    ehild

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    Of course.
     
  11. Feb 19, 2017 #10
    Ok, thanks, let me update the work and go from there.
     
  12. Feb 19, 2017 #11
    I started moving the [itex]\dfrac{dy}{dx}[/itex] to the right side instead, but didn't finish just yet.

    [itex]\dfrac{(x+y)(2x)-(\dfrac{dy}{dx})(x^2)}{(x+y^2)^2}=2y\cdot\dfrac{dy}{dx}[/itex]

    [itex]\dfrac{1}{\dfrac{dy}{dx}}\cdot\dfrac{2x^2+2xy-x^2\cdot\dfrac{dy}{dx}}{(x+y^2\cdot\dfrac{dy}{dx})^2}=2y\cdot\dfrac{dy}{dx}\cdot\dfrac{1}{\dfrac{dy}{dx}}[/itex]

    [itex]\dfrac{1}{2y}\cdot\dfrac{2x^2+2xy-x^2}{(x+y^2\cdot\dfrac{dy}{dx})^2}=\dfrac{2y\cdot\dfrac{dy}{dx}}{\dfrac{dy}{dx}}\cdot\dfrac{1}{2y}[/itex]

    [itex]\dfrac{2x^2+2xy-x^2}{2y(x+y\cdot\dfrac{dy}{dx})^2}=\dfrac{\dfrac{dy}{dx}}{\dfrac{dy}{dx}}[/itex]
     
  13. Feb 19, 2017 #12

    Ray Vickson

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    I am talking about the ##y \frac{dy}{dx}## part.
     
  14. Feb 19, 2017 #13
    That was my fault. It was supposed to be [itex](x+y)[/itex]
     
  15. Feb 19, 2017 #14

    ehild

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    Wrong denominator on the left side.
     
  16. Feb 19, 2017 #15

    ehild

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    You confuse yourself when dy/dx appears on both sides. It has no sense dividing by dy/dx.
    Multiply the original equation by (x+y), then you need to differentiate the equation ##x^2=(x+y)(y^2+8)##.
     
  17. Feb 19, 2017 #16
    I fixed that after that, it should've been like the rest.

    Are you saying that after getting rid of that (x+y) in the denominator, I would distribute on the right side and use the product rule?

    See this is my problem in math, I follow the pattern I first learn from when introduced into something new. When you said to multiply the original equation by (x+y), I wouldn't have thought of that because I've been following what I've been previously doing, by using the quotient rule because there's a fraction on the left side.

    I want to clarify on something, after doing this:

    [itex]x^2=(x+y)(y^2+8)[/itex]

    In the second step after distributing,

    [itex]x^2=xy^2\cdot\dfrac{dy}{dx}+8x+y^3\cdot\dfrac{dy}{dx}+8y\cdot\dfrac{dy}{dx}[/itex]

    When I get here, am I using the chain rule on xy^2 and the product rule on 8x and 8y?
     
  18. Feb 19, 2017 #17

    ehild

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    The rest was also wrong. There is no derivative in the denominator on the left side. Why did you put dy/dx there again? And I do not follow what you did.


    No need to distribute. Use the product rule, but differentiate both sides!
    It is all right to differentiate the original equation, but you get a more complicate equation for dy/dx, and you looked confused. You need to use basic rules of multiplication and addition, instead of "following patterns".
    the left side is x2. What is the derivative?
    The right side is (x+y)(y2+8). Apply product rule to get the derivative with respect x.
     
  19. Feb 19, 2017 #18
    My professor was saying whenever there's a "y", you need to multiply by [itex]\dfrac{dy}{dx}[/itex]. He never showed us an example of a problem given a fraction so I don't know about that.

    [itex]x^2=(x+y)(y^2+8)[/itex]

    [itex]f(x)=x+y[/itex]

    [itex]f'(x)=\dfrac{dy}{dx}[/itex]

    [itex]g(x)=y^2+8[/itex]

    [itex]g'(x)=2y\dfrac{dy}{dx}[/itex]

    [itex]x^2=(x+y)(2y\cdot\dfrac{dy}{dx})+(y^2+8)(\dfrac{dy}{dx})[/itex]

    [itex]x^2=2xy\cdot\dfrac{dy}{dx}+2y^2\cdot\dfrac{dy}{dx}+y^2\cdot\dfrac{dy}{dx}+8\cdot\dfrac{dy}{dx}[/itex]

    Would you distribute like I just did on the last part there? Doesn't seem correct.
     
  20. Feb 19, 2017 #19

    Mark44

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    You might have heard your professor incorrectly. When you're doing implicit differentiation, where there's a y, upon differentiation you get ##\frac{dy}{dx}##. You aren't multiplying by ##\frac{dy}{dx}##.

    For example, if there's a term of ##y^2##, differentiating it gives ##\frac d{dx}\left(y^2\right) = 2y\cdot \frac {dy}{dx}##. That last factor comes from the chain rule. ##\frac d{dx}\left(y^2\right) = \frac d {dy} \left(y^2\right) \cdot \frac {dy}{dx}##.
     
  21. Feb 19, 2017 #20

    ehild

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    You have to differentiate both sides! so the left side is d(x2)/dx = ?
    About the saying of your professor, whenever you differentiate a function of y, differentiate it with respect to y first, then multiply by y'. But in case of a fraction, it is f'g-fg' divided by the square of the original denominator.
     
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