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Homework Help: Differentiating for y

  1. Jul 23, 2011 #1
    differentiating for "y"

    1. The problem statement, all variables and given/known data
    I don't quite understand the reason why one can [itex]\frac{d}{dx}[/itex] y2 and get 2y [itex]\frac{dy}{dx}[/itex] when using implicit differentiation. Why can one even go as far as to change y2 to 2y?


    2. Relevant equations



    3. The attempt at a solution
    I see the chain rule is applied, I just would like to have a good understanding as to why differentiating y2 gives 2y [itex]\frac{dy}{dx}[/itex] when using implicit differentiation
     
    Last edited: Jul 23, 2011
  2. jcsd
  3. Jul 23, 2011 #2
    Re: differentiating for "y"

    you should think of it as y = f ( x ), as y is a function of x.
    Then, the use of the chain-rule is obvious. Suppose you have the function T ( x ) = 2 + [ f( x ) ]^2 = 2 + y^2 . Then if you want to differentiate this with respect to x, you have to differentiate T ' ( x ) = 2f(x )*f ' ( x )
     
  4. Jul 23, 2011 #3

    Ray Vickson

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    Re: differentiating for "y"

    For f(x) = [y(x)]^2 and small h > 0 we have f(x+h) - f(x) = [y(x+h)]^2 - [y(x)]^2. We also have [y(x+h) -y(x)]/h ~ y'(x) for small h > 0 (becoming exact in the limit h --> 0), so f(x+h) - f(X) is almost equal to [y(x) + h*y'(x)]^2 - [y(x)]^2 = 2*y(x)*y'(x)*h + h^2*y'(x)^2, with equality becoming exact in the limit. So, what is limit_{h->0} [f(x+h) - f(x)]/h?

    RGV
     
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