Implicit Differentiation

A function in which the dependent variable $y$ is expressed solely in terms of the independent variable $x$, i.e. $y=f(x)$, is called an explicit function — for instance $y=\tfrac12x^3-1$. An equivalent equation such as $2y-x^3+2=0$, which is not solved for $y$, is instead said to define $y$ implicitly, or to make $y$ an implicit function of $x$.

Why implicit functions matter. The equation $x^2+y^2=4$ describes a circle of radius $2$ centred at the origin. It is not itself a function, because for any $x$ with $-2<x<2$ there are two corresponding $y$-values: $f(x)=\sqrt{4-x^2}$ (the top half, $y\ge0$) and $g(x)=-\sqrt{4-x^2}$ (the bottom half, $y\le0$), for $-2\le x\le2$. Choosing either half individually does give a genuine function, so the single equation $x^2+y^2=4$ is said to define at least two distinct implicit functions of $x$ on $[-2,2]$; both $x^2+[f(x)]^2=4$ and $x^2+[g(x)]^2=4$ hold as identities on that interval.

In general, if $F(x,y)=0$ defines a function $f$ implicitly on some interval, then $F(x,f(x))=0$ is an identity on that interval, and the graph of $f$ is a portion (or all) of the graph of $F(x,y)=0$. A more complicated equation such as $x^4+x^2y^3-y^5=2x+1$ may determine several implicit functions on suitably restricted intervals, and it may not even be possible to solve algebraically for $y$ in terms of $x$ — yet the derivative $dy/dx$ can often still be found by a process called implicit differentiation.

The method. Differentiate both sides of the given equation with respect to $x$, treating $y$ throughout as a differentiable function of $x$ (so every term containing $y$ picks up a factor $\dfrac{dy}{dx}$ via the chain rule — most simply captured by the power-rule-for-functions form $\dfrac{d}{dx}(y^n) = ny^{n-1}\dfrac{dy}{dx}$, $n$ an integer), then solve the resulting equation algebraically for $\dfrac{dy}{dx}$.

Worked illustration. For $x^2+y^2=1$: differentiating termwise, $\dfrac{d}{dx}(x^2)+\dfrac{d}{dx}(y^2) = \dfrac{d}{dx}(1)$ gives $2x+2y\dfrac{dy}{dx}=0$, so $\dfrac{dy}{dx}=-\dfrac{x}{y}$ ($y\ne0$). Substituting $x=1$ into the original equation gives $y=\pm\sqrt{3}/\sqrt{4}$-style paired values (two points on two different implicit branches), and at each of those two points the same implicit formula $dy/dx=-x/y$ correctly reproduces the tangent slope — implicit differentiation handles both branches at once without ever having to solve for $y$ explicitly. …