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Mom's Basement University

Implicit Differentiation

What You’ll Learn

Section titled “What You’ll Learn”

In this lesson you’ll learn how to find dy/dx when y isn’t isolated on one side of the equation. This technique is called implicit differentiation, and it unlocks derivatives for curves like circles, ellipses, and other relationships where solving for y first would be messy or impossible.

The Concept

Section titled “The Concept”

So far, every function you’ve differentiated has been in the form y = f(x), where y is explicitly written in terms of x. But many important equations aren’t written that way.

Consider the equation of a circle:

x2+y2=25x^2 + y^2 = 25

You could solve for y (getting two square root expressions), but that’s awkward. Implicit differentiation lets you find dy/dx directly from the equation as written.

The process:

  1. Differentiate both sides with respect to x
  2. Every time you differentiate a term containing y, tack on a dy/dx (because y depends on x, so the chain rule applies)
  3. Solve for dy/dx

That’s it. The chain rule is doing all the work. When you see y² and differentiate with respect to x, the chain rule gives you 2y times dy/dx. When you see x², you just get 2x (no dy/dx needed because x is the variable you’re differentiating with respect to).

The blue circle is x² + y² = 25. You can’t write it as a single y = f(x) function (it’s two halves). But implicit differentiation gives you the slope at any point anyway. The green tangent line at (3, 4) has slope -3/4, found by implicit differentiation. The dashed purple line is the radius, which is always perpendicular to the tangent on a circle.

Worked Example

Section titled “Worked Example”

Example 1: Circle

Find dy/dx for x² + y² = 25.

Differentiate both sides with respect to x:

2x+2ydydx=02x + 2y\frac{dy}{dx} = 0

Solve for dy/dx:

2ydydx=−2x⇒dydx=−xy2y\frac{dy}{dx} = -2x \quad \Rightarrow \quad \frac{dy}{dx} = -\frac{x}{y}

At the point (3, 4) on the circle, the slope is -3/4. At (0, 5), the slope is 0 (horizontal tangent at the top). At (5, 0), the slope is undefined (vertical tangent on the side). All of that makes geometric sense if you picture the circle.

Example 2: Product of x and y

Find dy/dx for xy + y³ = 6.

The xy term needs the product rule (x and y are both functions of x):

(1⋅y+x⋅dydx)+3y2dydx=0\left(1 \cdot y + x \cdot \frac{dy}{dx}\right) + 3y^2\frac{dy}{dx} = 0

Group the dy/dx terms:

xdydx+3y2dydx=−yx\frac{dy}{dx} + 3y^2\frac{dy}{dx} = -y

Factor out dy/dx:

dydx(x+3y2)=−y⇒dydx=−yx+3y2\frac{dy}{dx}(x + 3y^2) = -y \quad \Rightarrow \quad \frac{dy}{dx} = \frac{-y}{x + 3y^2}

Example 3: Finding a tangent line

For x² + y² = 25, find the equation of the tangent line at (3, 4).

We already found dy/dx = -x/y. At (3, 4):

slope=−34\text{slope} = -\frac{3}{4}

Using point-slope form:

y−4=−34(x−3)y - 4 = -\frac{3}{4}(x - 3) y=−34x+94+4=−34x+254y = -\frac{3}{4}x + \frac{9}{4} + 4 = -\frac{3}{4}x + \frac{25}{4}

Real-World Application

Section titled “Real-World Application”

Implicit differentiation is essential when variables are naturally intertwined:

  • Physics: the ideal gas law PV = nRT relates pressure, volume, and temperature. If you want dP/dT while V changes, implicit differentiation handles it.
  • Economics: indifference curves and production possibility frontiers are defined implicitly. Implicit differentiation gives marginal rates of substitution.
  • Engineering: stress-strain relationships in materials are often implicit. Finding how one changes with respect to the other requires this technique.
  • Geometry: finding tangent lines to circles, ellipses, and other curves defined by equations.

Quiz

Section titled “Quiz”
When differentiating y² with respect to x, you get:
For x² + y² = 25, dy/dx equals:
Implicit differentiation is needed when:
When differentiating xy with respect to x, you use:
At the point (0, 5) on x² + y² = 25, the tangent line is: