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Implicit Differentiation

  1. Jun 6, 2012 #1
    I am reading about this topic, and I came across this sentence "Remember, every time we want to differentiate a function of y with respect to x, we differentiate with respect to y and then multiply by dy/dx." What exactly does this mean?
     
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  3. Jun 6, 2012 #2
    Basically, what we're doing here is use the chain rule. You are trying to find

    [tex]\frac{d}{dx}f(y)[/tex]

    Which after the chain rule becomes,

    [tex]\frac{df(y)}{dx} = \frac{df(y)}{dy} \cdot \frac{dy}{dx}[/tex]
     
  4. Jun 6, 2012 #3
    So, does it essentially say that first find how y changes with itself, and then find how it changes with x?

    For instance, [itex]\frac{d}{dy}[y^2][/itex] Which would equal [itex]2y\frac{dy}{dx}[/itex] Where the 2y is how y changes with respect to itself, and the other part is how y changes with x?
     
  5. Jun 6, 2012 #4
    An excerpt from my calculus books reads, "...when you differentiate terms involving y, you must first apply the chain rule, because you are assuming that y is defined implicitly as a differentiable function of x." Why are we allowed to assume in this case? And what does it mean to assume this?

    EDIT:

    I just read on a website that a function that is implicit not a function that is not depend on x, but it's just not exactly clear how it is dependent upon x. Is that why we can assume, because even though it is not clear how y is depending on x, it still is, we just can't show exactly how it depends on x?
     
    Last edited: Jun 6, 2012
  6. Jun 6, 2012 #5

    pcm

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    that is equal to 2y because you differentiate wrt y.
     
  7. Jun 6, 2012 #6

    HallsofIvy

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    What you have written is not correct, but may just be a typo. You meant to write [itex]\frac{d}{dx}[y^2][/itex], not [itex]\frac{d}{dy}[y^2][/itex].
     
  8. Jun 6, 2012 #7

    AlephZero

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    If you write an equation that shows "exactly" how y depends on x, that doesn't mean you can turn it into an equation that says "y = a formula that only contains x".

    Try it yourself with an equation like ##\sin(x/y) + \cos(y/x) = 0## for example.

    An "implicit function" is where the x's and y's mixed up together. An "explicit function" has the x's and y's separated out from each other.
     
    Last edited: Jun 6, 2012
  9. Jun 6, 2012 #8
    HallsofIvy, could you possible explain the procedure of the example problem given in the link?

    http://www.mash.dept.shef.ac.uk/Resources/web-implicit.pdf

    Page 4 section 3.

    In the example problem, I know they ultimately want to ascertain how the function changes with x--or, in other words, take the derivative with respect to x. But you can see that the first take the derivative with respect to y for the y terms and then multiply by dy/dx. Actually, in their steps to implicit differentiation, they state that, "to differentiate a function of y with respect to x, we differentiate with respect to y and then multiply by dy/dx. I don't understand the point in differentiating with respect to y first, then multiplying by dy/dx.
     
  10. Jun 7, 2012 #9
    Ok so the first step they are telling you they're taking the derivative wrt to x.
    The derivative of y2 wrt to x is 2yy'. The derivative of 3x2 is 6x. d/dx of y3 is 3y2y'. And finally, d/dx of 3y is just 3y'.

    Then they rearranged the equation so that the y' are on one side. They then factor out a y' and divide by the stuff that's left.
     
  11. Jun 8, 2012 #10
    As said before, it is because of the chain rule.

    When you have a function which depends on y, f = f(y), where y depends on x, y = y(x), the derivative df/dx can be multiplied by dy/dy (which is 1 and so does not change the value) and reorganized:

    [tex]\frac{df}{dx} = \frac{df}{dx}\frac{dy}{dy} = \frac{df}{dy}\frac{dy}{dx}[/tex]

    This is used in the example to find the dependance that y has on x, given an equation of only y, x and constants.
     
  12. Jun 8, 2012 #11
    Okay, I am fairly certain I have just one more question. In an article, they explain implicit differentiation by means of an example. The problem is [itex]x^2 + y^2 = 10[/itex], which they rewrite as [itex]x^2 + [f(x)]^2 = 10[/itex] just so that it is clear that y is still a function of x, just not explicitly. One thing they say that confuses me is, "We can think about this equation as the equality between two functions of x. If these two functions are equal, their derivatives must be equal too. Just common sense." What do they mean by this?
     
  13. Jun 8, 2012 #12

    HallsofIvy

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    Yes, that's true. the two functions of x involved are [itex]x^2+ [f(x)]^2[/itex] and 10. Since they are equal, they must have the same derivative. And since the derivative of the constant function, 10, is 0, the derivative of [itex]x^2+ [f(x)]^2[/itex], with respect to x, must be 0. So from [itex]x^2+ y^2= 10[/itex], differentiating both sides with respect to x, we have [itex]2x+ 2y(dy/dx)= 0[/itex]. Now, you can solve that for dy/dx.
     
  14. Jun 8, 2012 #13
    Oh, yes, very interesting. Thank you everyone.
     
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