Turn on the graph of the derivative and change the value of a with the slider. You can see that the line intersects the red circle at the points where the gradient of the curve is equal to "a" (the value of the derivative). This is probably best seen by setting a=0, 1 or -1.
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Below is the expression for the derivative when calculated using the quotient rule on the second equation. The one that is undefined for (0, 0).
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If you turn on that graph and change the value of a with the slider you can see the orange lines again intersect the circle at the points where the gradient is equal to "a". This shows that the second algebraic form of the derivative gives the same values for the derivative as the first form for all points on the circle.
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The difference this time is that the derivative appears to intersect the circle at (0, 0) but of course it doesn't as the circle doesn't exist at at (0, 0) if we're using the second equation.
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For fun I have included a third form of the derivative calculated using the quotient rule on a different algebraic form for the circle. The equation and derivative expression are given below.
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Turn on both of those graphs and see how the graph for the derivative and the circle intersect in the way we have seen previously. This shows that once again the third algebraic form for the derivative gives the same values for the gradient of the circle as the first even though the graph of the derivative itself looks a little crazy. Again both sides of the equation for the circle are undefined for (0, 0).
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So it is possible to get different algebraic forms for the derivative of an implicitly defined curve but these different forms will give the same value for dy/dx for all points on the curve. They will give different values for dy/dx if you substitute points that are not on the curve which is why the expressions are algebraically different and look very different graphically.
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