3.8 Implicit Differentiation

Implicit differentiation lets you find derivatives when you can't easily isolate $y$ as a function of $x$. Many important curves, like circles and ellipses, are defined by equations where $x$ and $y$ are mixed together, and this technique handles them cleanly.

You'll use implicit differentiation constantly in related rates problems and whenever you need the slope of a tangent line to a curve that isn't written as $y = f(x)$. The core idea: differentiate both sides of the equation with respect to $x$, apply the chain rule to any $y$ terms, and then solve for $\frac{dy}{dx}$.

Implicit Differentiation

Implicit differentiation for complex functions

An explicit function spells out $y$ directly, like $y = x^2 + 3$. An implicit equation mixes $x$ and $y$ together, like $x^2 + y^2 = 25$. You can't always solve for $y$ neatly, so you need a different approach.

The key insight is that $y$ is still a function of $x$, even if you can't write it out explicitly. So every time you differentiate a term containing $y$, you must apply the chain rule and tack on a $\frac{dy}{dx}$.

Steps for implicit differentiation:

1. Differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$.
2. Every time you differentiate a $y$-term, multiply by $\frac{dy}{dx}$ (this is the chain rule in action).
3. Collect all terms with $\frac{dy}{dx}$ on one side of the equation.
4. Factor out $\frac{dy}{dx}$ and solve.

Example 1: $x^2 + y^2 = 25$

Differentiate both sides:

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

Solve for $\frac{dy}{dx}$:

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

Notice the derivative depends on both $x$ and $y$. That's normal with implicit differentiation.

Example 2: $x^3 + y^3 = 6xy$

Differentiate both sides. The right side requires the product rule:

$3x^2 + 3y^2\frac{dy}{dx} = 6y + 6x\frac{dy}{dx}$

Collect $\frac{dy}{dx}$ terms on one side:

$3y^2\frac{dy}{dx} - 6x\frac{dy}{dx} = 6y - 3x^2$

Factor and solve:

$\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x}$

You'll frequently need the product rule, quotient rule, or both when the equation has terms like $xy$, $\frac{x}{y}$, or similar combinations.

Tangent slopes of implicit curves

The slope of the tangent line at any point still equals the derivative at that point. The only difference is that you find $\frac{dy}{dx}$ through implicit differentiation, then plug in the specific coordinates.

Steps:

1. Use implicit differentiation to find $\frac{dy}{dx}$ (which will typically involve both $x$ and $y$).
2. Substitute the coordinates of the given point into your expression for $\frac{dy}{dx}$.

Example 1: Find the slope of the tangent line to $x^2 + y^2 = 25$ at the point $(3, 4)$.

From above, $\frac{dy}{dx} = -\frac{x}{y}$. Substituting:

$\frac{dy}{dx}\bigg|_{(3,4)} = -\frac{3}{4}$

The tangent line at $(3, 4)$ has slope $-\frac{3}{4}$.

Example 2: Find the slope of the tangent line to $xy = 10$ at the point $(2, 5)$.

Differentiate using the product rule:

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

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

Substitute $(2, 5)$:

$\frac{dy}{dx}\bigg|_{(2,5)} = -\frac{5}{2}$

Related rates via implicit differentiation

Related rates problems give you a relationship between quantities that are all changing over time, plus the rate of change of one quantity, and ask you to find the rate of change of another. Implicit differentiation is the engine that connects these rates, except now you differentiate with respect to $t$ (time) instead of $x$.

Steps to solve related rates problems:

1. Draw a picture and label all quantities that change with variables (not numbers yet, unless a quantity is constant).
2. Write an equation that relates the relevant quantities.
3. Differentiate both sides with respect to $t$, applying the chain rule to every variable.
4. Substitute the known values after differentiating (not before).
5. Solve for the desired rate.

Example 1: Sliding ladder

A 13 ft ladder leans against a wall. The bottom slides away at 2 ft/sec. How fast is the top sliding down when the bottom is 5 ft from the wall?

The Pythagorean theorem relates the quantities:

$x^2 + y^2 = 169$

where $x$ is the distance from the wall and $y$ is the height on the wall. Differentiate with respect to $t$:

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

When $x = 5$, find $y$: $y = \sqrt{169 - 25} = 12$.

Substitute $x = 5$, $y = 12$, and $\frac{dx}{dt} = 2$:

$2(5)(2) + 2(12)\frac{dy}{dt} = 0$

$20 + 24\frac{dy}{dt} = 0$

$\frac{dy}{dt} = -\frac{20}{24} = -\frac{5}{6} \text{ ft/sec}$

The negative sign means the top is sliding down at $\frac{5}{6}$ ft/sec.

Example 2: Filling a cylindrical pool

A cylindrical pool with a fixed radius $r$ is being filled at 5 ft³/min. Find the rate at which the water level rises when the water is 3 ft deep.

Since the radius is constant (the pool doesn't change shape), the volume formula simplifies:

$V = \pi r^2 h$

Differentiate with respect to $t$. Because $r$ is constant, only $h$ changes:

$\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}$

Substitute $\frac{dV}{dt} = 5$ and solve:

$\frac{dh}{dt} = \frac{5}{\pi r^2}$

Notice the answer depends on the pool's radius. If $r = 3$ ft, for instance, $\frac{dh}{dt} = \frac{5}{9\pi} \approx 0.177$ ft/min. The depth of 3 ft doesn't actually matter here because a cylinder has the same cross-section at every height.