   SEARCH HOME Math Central Quandaries & Queries  Question from Pranay, a student: Is a circle $x^2+y^2=2$ a function? If it is not a function, why is it possible to do implicit differentiation on it? Thanks. Hi,

$x^2 + y^2 = 2$ does not define a function, in fact if you solve this equation for $y$ you get

$y = \pm \sqrt{2 - x^2}$

which you can write as two functions

$y = \sqrt{2 - x^2} \mbox{ and } y = - \sqrt{2 - x^2}.$

I can differentiate each of these functions.

If $y = \sqrt{2 - x^2}.$

$\frac{dy}{dx} = \frac12 \left(2 - x^2\right)^{-1/2} (-2x) = \frac{-x}{\left(2 - x^2\right)^{1/2}}$

and since $y = \sqrt{2 - x^2}$ this can be written

$\frac{dy}{dx} = \frac{-x}{y}.$

If $y = -\sqrt{2 - x^2}.$

$\frac{dy}{dx} = -\frac12 \left(2 - x^2\right)^{-1/2} (-2x) = \frac{x}{\left(2 - x^2\right)^{1/2}}$

and since $y = -\sqrt{2 - x^2}$ this can be written

$\frac{dy}{dx} = \frac{-x}{y}.$

Notice that in differentiating each of the two functions I used the chain rule.

Implicit differentiation allows us to show that

$\frac{dy}{dx} = \frac{-x}{y}$

without solving for $y.$ What we can do is think of $y$ as a function of $x,$ even though we don't know an explicit expression for $y.$

Differentiate both sides of $x^2 + y^2 = 2$ with respect to $x.$ The right side is a constant so its derivative is zero. Since $y$ is some function of $x$ I will have to use the chain rule to differentiate it. Thus I get

$2x + 2 y \frac{dy}{dx} = 0$

and simplifying I get

$\frac{dy}{dx} = \frac{-x}{y}.$

The real power of implicit differentiation is seen if you can't solve for $y$ as we did above. For example if

$x \; y^7 - \frac{x}{y^2} = x^3 y$

then implicit differentiation allows you to find an expression for $\large \frac{dy}{dx}$ in terms of $x$ and $y.$

I hope this helps,
Penny      Math Central is supported by the University of Regina and the Imperial Oil Foundation.