Concept preview

What you will be able to do

1. 01Explain why, when $y$ is defined implicitly as a function of $x$, differentiating any $y$-term produces a factor of $\dfrac{dy}{dx}$ by the chain rule.
2. 02Differentiate both sides of an implicit equation with respect to $x$, collect the $\dfrac{dy}{dx}$ terms, and solve algebraically for $\dfrac{dy}{dx}$.
3. 03Use implicit differentiation to find the slope of a tangent line to a curve that is not solved for $y$ — for the circle $x^2+y^2=25$ obtaining $\dfrac{dy}{dx}=-\dfrac{x}{y}$ and a slope of $-\dfrac{3}{4}$ at the point $(3,4)$.

In implicit differentiation we assume the equation secretly defines $y$ as one or more functions of $x$, even if we never write those functions down. Because $y$ is then a function of $x$, any expression built from $y$ is a composite — a function of $y$, which is itself a function of $x$. Differentiating such a composite with respect to $x$ demands the chain rule. For instance, $\dfrac{d}{dx}\bigl(y^2\bigr)$ is not $2y$; the outer function "square" gives $2y$, and then the chain rule multiplies by the derivative of the inside, $y$, which is $\dfrac{dy}{dx}$. So $\dfrac{d}{dx}\bigl(y^2\bigr)=2y\cdot\dfrac{dy}{dx}.$ Every $y$-term carries this extra $\dfrac{dy}{dx}$ tag; every pure-$x$ term does not, because the derivative of $x$ with respect to $x$ is just $1$.