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    Why does implicit differentiation work? That is why is  \frac{d}{dx} (f(y)) = \frac{d}{dy} (f(y)) \times \frac{dy}{dx}?

    And this
    http://en.wikipedia.org/wiki/Vertical_line_test
    Does that mean that  y = x^2 is a function but  x = y^2 and  |y| = \sin x are not?
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    (Original post by gangsta316)
    Why does implicit differentiation work? That is why is  \frac{d}{dx} (f(y)) = \frac{d}{dy} (f(y)) \times \frac{dy}{dx}?
    Because the dy terms cancel. >_><_<>_>

    http://en.wikipedia.org/wiki/Vertical_line_test
    Does that mean that  y = x^2 is a function but  x = y^2 and  |y| = \sin x are not?
    Correct. Well, they're not functions of y, any way.
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    1. Consider the chain rule.

    2. They are not functions (where the definition of function requires one-one mapping).
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    1) It's a result of the chain rule. Write

    z = f(y)

    Now we're trying to find dz/dx, which we know is

    dz/dx = dz/dy * dy/dx (chain rule)

    But dz/dy = f'(y), so

    dz/dx = f'(y) * dy/dx, as required.

    2) Haven't read the article but y = x^2 is a many-to-one function (i.e. many values can lead to the same answer, e.g. (-5)^2 = 5^2 = 25). x = y^2 is not a function of y and neither is the other one.

    If one value of x can lead to more than one value of y, then it's not a function.
 
 
 
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