When y is hiding inside an equation

Implicit differentiation and inverse slopes are the same idea: related rates of x and y.

Learning goals

Differentiate an implicit relation and solve for dy/dx.
Find an inverse function’s derivative at a point from the original slope.

Explanation

If x² + y² = 25, y is still a function of x on a chosen branch. Differentiate both sides with respect to x, treating y as y(x). You get 2x + 2y y′ = 0, so y′ = −x/y where y ≠ 0. That is the tangent slope to the circle.

If g is the inverse of f, the graphs swap x and y, so slopes are reciprocals at matching points. If f(2) = 5 and f′(2) = 4, then g′(5) = 1/4. You do not always need an explicit formula for g.

Key terms

Implicit differentiation. Differentiating both sides of an equation in x and y while treating y as a function of x.
Inverse function derivative. At matching points, the inverse’s slope is the reciprocal of the original function’s slope.

Common mistakes

Differentiating y² as 2y and dropping y′.
Reciprocating f′ at the wrong x-value instead of at the paired point.

Practice

Original Marlow Works items. Check the answer explanation after you try.

1 of 2Limits and Continuity · easy · multiple choice

What is lim (x→3) of (x² − 9) / (x − 3)?

Original Marlow Works item — not a College Board question.

Take your time—this is practice, not a test.

Answer explanation

f(3) = 7 and f′(3) = 5. If g = f⁻¹, what is g′(7)?

Answer. 1/5.

The inverse slope at 7 is the reciprocal of the original slope at 3.

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Related resources

External links with reuse status. Marlow Works is independent and does not copy restricted exam or textbook material.

Reference only — do not copy
OpenStax College Algebra 2e
Algebra reference; current terms include noncommercial and AI-ingestion limits.
OpenStax · Restricted — see publisher page · accessed 2026-10-01
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