3.2 Implicit Differentiation

What is implicit differentiation?

When y can't easily be solved for in terms of x, we differentiate both sides of the equation with respect to x, treating y as a function of x. Every time you differentiate a term with y, apply the chain rule and multiply by dy/dx. Then solve algebraically for dy/dx.

Key mechanics

Differentiate term-by-term using normal rules (power, product, quotient). For a term like y², get 2y(dy/dx). For xy, use the product rule: 1·y + x(dy/dx). After differentiating everything, collect all dy/dx terms on one side, factor out dy/dx, and divide to isolate it.

Evaluating at a point

To find the slope of a tangent line at a specific point, plug the point's x and y values directly into the dy/dx expression (no need to fully simplify first). This is common for finding tangent lines to circles, ellipses, or other curves not written as y = f(x).

Key terms

Implicit differentiation

A technique to find dy/dx when y is not isolated, by differentiating both sides with respect to x and treating y as a function of x.

Differentiating y² with respect to x

2y(dy/dx), using the chain rule since y is a function of x.

Differentiating xy with respect to x

y + x(dy/dx), using the product rule.

Steps to solve for dy/dx

1) Differentiate both sides term-by-term. 2) Collect dy/dx terms on one side. 3) Factor out dy/dx. 4) Divide to isolate dy/dx.

Finding slope at a point implicitly

Differentiate implicitly to get dy/dx, then substitute the given x and y values into the expression.

Try an AP-style question

For x² + y² = 25, what is the slope of the tangent line at (3, 4)?

1. −4/3
2. 4/3
3. −3/4
4. 3/4