# Implicit Differentiation Misunderstanding.

Say we are given the unit circle, defined by all real numbers $x,y$ such that $x^2+y^2=1$. Now, if one tried to solve explicitly for $y$, one would get that $y=\pm{\sqrt{1-x^2}}$ and this is obviously not a function, so traditional calculus techniques cannot be used. As $x^2+y^2=1$ defines our curve, we can take the derivative of all terms with respect to $x$ to get that $2x+2y\frac{dy}{dx}=0$, as by the chain rule $\frac{d(y^2)}{dx}=\frac{d(y^2)}{dy}\frac{dy}{dx}$, and one can therefore do some algebraic manipulation to get that $\frac{dy}{dx}=-\frac{x}{y}$. However, I have a few questions:

1) I thought the whole point was that $y$ is not a function, and so $\frac{dy}{dx}$ does not make sense as this notation is defined only when $y$ is a function (or am I wrong in this regard?)

2) Why are we allowed to take the derivative with respect to $x$ on both sides of the equation, but not for example to solve $x^2=1$ to get $2x=0$ which would produce the wrong answer that $x=0$ (my inkling tells me that this is because $x^2=1$ is not an identity but an equation, but I just want to check up on this...)

• $y=\pm{\sqrt{1-x^2}}$ is a collection of 2 functions. – Love Invariants Aug 1 '18 at 18:14
• @LoveInvariants Therefore.... and what about for more complex implicit curves – Daniele1234 Aug 1 '18 at 18:14
• They too are are collection of functions because function is defined only for linear $y$ i.e. one value of $y$ for a value of $x$ – Love Invariants Aug 1 '18 at 18:17
• $y=\pm{\sqrt{1-x^2}}$ isn't a function but $y=\sqrt{1-x^2}$ and $y=-\sqrt{1-x^2}$ are functions. – Love Invariants Aug 1 '18 at 18:27

Good question!

1) You're right that the equation $x^2+y^2=1$ does not define $y$ as a function of $x$ globally. However, if you zoom in on certain points of the curve, what you see does look like the graph of a function. Because zooming in will allow you to ignore that “other” branch of the circle.

The fancy way to say this is that for most points on the curve, we can find a neighborhood of the point for which the intersection of the curve and the neighborhood is the graph of a function. And the derivative of that function is what's predicted by the implicit differentiation process.

But not necessarily every point on the curve. For instance, on the circle near $(-1,0)$, no amount of zooming in will make that look the like the graph of a function. And it's no coincidence that the tangent line is vertical there.

I have some slides with pictures up on this topic from several years ago.

2) The equations $x^2+y^2=1$ and $x^2=1$ are much different than they appear. The first equation relates two variables $x$ and $y$: for any value of $x$ we can find (“solve for”) all related values of $y$.

The second equation is not a relation among variables; there's only one variable after all. It's an equation describing specific, unknown numbers $x$. We can only differentiate functions (or implicitly differentiate certain relations). Since $x^2=1$ is not an equation of functions, differentiating doesn't apply. To give an even more ludicrous example: Suppose we have the equation $2x=4$. If we differentiate both sides, we get $2=0$, a contradiction, which would indicate there was no solution at all. Of course this is wrong since $x=2$ satisfies the equation.

• Thank you for your answer and powerpoint. So why is it exactly that you can't take the derivative of both sides for $x^2=1$, however you can for an identity of functions. – Daniele1234 Aug 1 '18 at 19:10
• @Daniele1234: I expanded on the second point; hope it helps. If not, you might consider asking a separate question about it. It probably deserves to be answered on its own. – Matthew Leingang Aug 1 '18 at 19:45

To be differentiable, an object doesn’t have to strictly be a function. There are a large family of curves and other objects which are not functions but can be approximated by tangent lines(or planes, etc).

As for your second question, it is not true in general that if we have an equality in an equation, this will hold true when the equation is differentiated. (This is due to the fact that differentiation ‘gets rid’ of information one can say).

• What do you mean by "differentiation gets rid of information" - that is, why can't one take the derivative of both sides of an equation – Daniele1234 Aug 1 '18 at 19:11
• @Daniele1234 $\frac{\text d}{\text dx}[f(x)+C] = \frac{\text d}{\text dx}[f(x)]$ for any constant $C$. In other words, $f'(x) = g'(x)$ does not necessarily mean that $f(x) = g(x)$. Thus, some distinguishing information is lost in the process of differentiation. – AlexanderJ93 Aug 1 '18 at 19:51