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My question is simple.

We are given the following function: $\frac{x^2-y^2}{x^2+y^2}=\frac{1}{2} $.

We are asked to find the derivative implicitly.

If we use the chain rule on the left-hand side, we can solve for $\frac{dy}{dx} = \frac{y}{x}$.

But if we multiply out the denominator first and then take the derivative, we end up with a different result, namely $\frac{dy}{dx}=\frac{x}{3y}$.

What is the reason? What is the theory behind this duplicity?

dacastr
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    The original equation can be rewritten as $2x^2-2y^2=x^2+y^2\implies x^2=3y^2$ so, really, your equations are equivalent. – lulu Nov 06 '16 at 23:27
  • But I find this just as strange and mysterious, @lulu. If I set my derivatives equal to each other, I get back the original function. How is this? Why is this happening? Any thoughts? – dacastr Nov 06 '16 at 23:44
  • You're not 'setting' your derivatives equal to each other. They're the same derivative. The reason you get your original equation back is because you've used the original equation to transform one expression into the other. – Sam Weatherhog Nov 07 '16 at 00:37
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    A better way of putting this is that the two expressions you have for the derivative are always equal on your curve. – Sam Weatherhog Nov 07 '16 at 00:45

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