Implicit Differentiation
Differentiate equations where y is defined implicitly as a function of x. Apply the chain rule to terms involving y (multiplying by dy/dx) and solve for dy/dx. Find slopes of tangent lines to circles, ellipses, and implicit curves.
A lesson of Calculus, in The Chain Rule and Advanced Differentiation, about 10 minutes.
By the end of this lesson, you will be able to differentiate implicitly defined functions by treating $y$ as a function of $x$, apply the Chain Rule whenever $y$ appears, find $dy/dx$ for equations not solvable for $y$, and find tangent lines to implicit curves.
The Big Idea
Not every curve can be written as $y = f(x)$ — circles, ellipses, hyperbolas, foliums. Implicit differentiation differentiates both sides of the equation with respect to $x$, treating $y$ as a function of $x$ (chain rule whenever $y$ appears), then solves for $dy/dx$.
Rule: whenever you differentiate a term involving $y$, multiply by $dy/dx$.
$$\frac{d}{dx}[y^n] = ny^{n-1}\frac{dy}{dx}, \quad \frac{d}{dx}[\sin y] = \cos y\frac{dy}{dx}, \quad \frac{d}{dx}[e^y] = e^y\frac{dy}{dx}$$
Example 1 — Circle
$x^2 + y^2 = 25$. Find $dy/dx$.
Differentiate both sides:
$2x + 2y\dfrac{dy}{dx} = 0$
$\dfrac{dy}{dx} = -\dfrac{x}{y}$
Verify: at $(3, 4)$: $dy/dx = -3/4$. This is the slope of the tangent to the circle at $(3,4)$. Explicit approach: $y = \sqrt{25-x^2}$, $y' = -x/\sqrt{25-x^2} = -3/4$ at $x = 3$ ✓.
Example 2 — Ellipse
$9x^2 + 16y^2 = 144$. Find $dy/dx$ at $(2, 3)$.
$18x + 32y\dfrac{dy}{dx} = 0 \implies \dfrac{dy}{dx} = -\dfrac{18x}{32y} = -\dfrac{9x}{16y}$
At $(2, 3)$: $dy/dx = -18/48 = -3/8$
Example 3 — Mixed Implicit
$xy + y^3 = 5$. Find $dy/dx$.
$y + x\dfrac{dy}{dx} + 3y^2\dfrac{dy}{dx} = 0$
$\dfrac{dy}{dx}(x + 3y^2) = -y$
$\dfrac{dy}{dx} = \dfrac{-y}{x + 3y^2}$
Example 4 — Trig Implicit
$\sin(xy) = x$. Find $dy/dx$.
$\cos(xy)\left(y + x\dfrac{dy}{dx}\right) = 1$
$x\cos(xy)\dfrac{dy}{dx} = 1 - y\cos(xy)$
$\dfrac{dy}{dx} = \dfrac{1 - y\cos(xy)}{x\cos(xy)}$
Example 5 — Second Implicit Derivative
$y^2 = x^3$. Find $y''$.
First: $2y\cdot y' = 3x^2 \implies y' = \dfrac{3x^2}{2y}$
Second: differentiate $2yy' = 3x^2$ again:
$2(y')^2 + 2yy'' = 6x$
$y'' = \dfrac{6x - 2(y')^2}{2y}$
Substitute $y' = 3x^2/(2y)$:
$y'' = \dfrac{6x - 2(9x^4/4y^2)}{2y} = \dfrac{6x - 9x^4/(2y^2)}{2y} = \dfrac{12xy^2 - 9x^4}{4y^3}$