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D/dx y^2 help

  1. Jun 29, 2013 #1
    Hi all!
    In questions having implicit functions,
    this expression -> "d/dx y^2" often appears in the calculation process.
    I use the chain rule to convert it to 2y x dy/dx
    But why can the chain rule be used here?
    I actually don't understand at all..
    Any help is appreciated.
    Thanks guys!
  2. jcsd
  3. Jun 29, 2013 #2
    your calculation is correct? what is the problem?

    okay so, lets lets set ##f = y ^2## this means if you take the gradient of it you get ## df = f_x dx + f_y dy## this is the chain rule, no? since the derivative of x is zero then you can say ## f_y = 2y## plugging this in you get ## df = 2y dy ## since we have established that x was zero, but to get your answer all you have to do is "take" the rate of change of x, since that is the parameter you wish to "constrain" your self to and get ## \frac{df}{dx} = 2y \frac{dy}{dx}##
  4. Jun 29, 2013 #3


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    This is NOT correct! (Unless that middle "x" indicates multiplication- a very bad notation when x is also used as a variable!) The derivative "d/dx y^2" is 2y dy/dx.

    It's pretty straight forward. [itex]u(y)= y^2[/itex] where y(x) is a function of x.
    The chain rule says that
    [tex]\frac{du}{dx}= \frac{du}{dy}\frac{dy}{dx}[/tex]

    Or course, since [itex]u= y^2[/itex], [itex]du/dy= 2y[/itex] so that
    [tex]\frac{du}{dx}= \frac{du}{dy}\frac{dy}{dx}= 2y \frac{dy}{dx}[/tex]

    It's that simple.
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