Implicit Differentiation: Method, Steps and Examples

Implicit differentiation lets you find the slope of a curve even when y cannot be isolated on one side of the equation. For a circle like x² + y² = 25, solving for y gives two messy branches with square roots — but differentiating the equation as it stands, treating y as a function of x, gives the slope directly in one clean step.

This matters because most real relationships between variables are not written as y = something. Circles, ellipses, the folium of Descartes, and countless physics equations (like the ideal gas law relating P, V, and T) are implicit. Without this technique, finding a tangent to any of them would be painful or impossible.

Key Concepts

• Implicit function: an equation F(x, y) = 0 where y is understood to depend on x, even though it is not solved for y.
• Chain rule on y-terms: when differentiating a term containing y, apply the chain rule: d/dx(y²) = 2y · dy/dx, and d/dx(xy) = y + x · dy/dx by the product rule.
• Notation: dy/dx is often written as y' to keep the algebra compact while solving.
• Result form: the derivative usually ends up expressed in terms of both x and y, which is perfectly fine — you substitute the point's coordinates to get a number.

How It Works

1. Step 1: Differentiate every term of the equation with respect to x, remembering that y is a function of x.
2. Step 2: For each y-term, multiply by dy/dx (the chain rule step).
3. Step 3: Collect all terms containing dy/dx on one side of the equation.
4. Step 4: Factor out dy/dx and divide to solve for it.
5. Step 5: Substitute the point's x and y values to find the numerical slope.

Worked Example

Find the slope of the tangent to the curve x³ + y³ = 6xy at the point (3, 3).

1. Step 1: Differentiate term by term: d/dx(x³) = 3x²; d/dx(y³) = 3y² y'; d/dx(6xy) = 6y + 6x y' (product rule).
2. Step 2: The differentiated equation is 3x² + 3y² y' = 6y + 6x y'.
3. Step 3: Collect y' terms on the left: 3y² y' − 6x y' = 6y − 3x².
4. Step 4: Factor: y'(3y² − 6x) = 6y − 3x², so y' = (6y − 3x²) / (3y² − 6x), which simplifies to (2y − x²) / (y² − 2x).
5. Step 5: At (3, 3): y' = (6 − 9) / (9 − 6) = (−3)/3 = −1. The tangent at (3, 3) has slope −1.

Common Misconceptions

• "Differentiating y² gives 2y." It gives 2y · dy/dx — the chain rule factor is the whole point, and forgetting it is the number one error.
• "I must solve for y first." That defeats the purpose. The technique exists precisely so you never have to isolate y.
• "The derivative should only contain x." An implicit derivative normally contains both x and y. That is expected and correct; you just plug in both coordinates.

Real-World Applications

• Physics: the ideal gas law PV = nRT relates three variables; implicit differentiation gives how pressure changes with volume at constant temperature.
• Economics: indifference curves and production functions are implicit relations, and their slopes (marginal rates of substitution) come from implicit differentiation.
• Engineering: stress-strain and load-deflection relationships are often implicit, and tangent slopes give stiffness values.
• Related rates: almost every related-rates problem starts by implicitly differentiating an equation like V = (4/3)πr³ with respect to time.

To recap: differentiate everything with respect to x, attach dy/dx to every y-term via the chain rule, collect, and solve. The slope at any point then follows by substitution.