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Suppose I wish to compute the implicit derivative of $\sqrt{x^2+y^2}=x+y$. One could differentiate both sides with respect to $x$, yielding $y\prime$ which we can make the subject.

Say I were to first square both sides, getting the expression $x^2+y^2=(x+y)^2$, and then I were to proceed as usual.

My question is, is this a valid method of computing the implicit derivative? Can squaring both sides cause problems?

Trogdor
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    Square both sides and look at the result. – By the way: There is no such thing as taking the implicit derivative of an equation. You can, however, find the derivative of some implicitly defined function $f$ without first obtaining an explicit expression for $f$. – Christian Blatter Jan 13 '15 at 12:12

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In general, it can cause minor problems, since the squared equation isn't the same as the original. The simplest example:

$$x = y$$ isn't the same as
$$x^2 = y^2$$ since the latter has solutions $$x = \pm y$$

In the equation in your question there isn't a problem because the LHS is never negative, assuming that you're following the usual convention that $\sqrt{x}$ denotes the non-negative square root of $x$.

However, that equation is a degenerate hyperbola, so its derivative is a bit funny anyway.

PM 2Ring
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  • For the example that you provided, could you not do this?

    $y\prime=x/y=1$ since $y=x$ by definition, which matches conventional differentiation of $y=x$

     – Trogdor Jan 13 '15 at 12:40
  • @Trogdor: Ah, but the squared equation in my example also has y' = -1, which is not correct for the original equation. – PM 2Ring Jan 13 '15 at 12:50
  • The squared equation may yield $y\prime=-1$, had we used that purely, but wouldn't it make more sense to use the original function $y=x$, instead of the 'tampered' one? – Trogdor Jan 13 '15 at 12:54
  • @Trogdor: Certainly! – PM 2Ring Jan 13 '15 at 13:02