Chain Rule in Implicit Differentiation: Step-by-Step Guide

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By Bikash Roy

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Bikash Roy

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· June 1, 2026

What Is Implicit Differentiation?

Most functions are explicit: y is given directly in terms of x (y = x² + 3x). In implicit functions, x and y are mixed in a single equation that cannot be easily solved for y — or would produce multiple branches if solved.

Examples of implicit equations:

• x² + y² = 25 (circle — two y solutions for each x)
• x³ + y³ = 6xy (folium of Descartes)
• e^(xy) = x + y
• sin(x + y) = y²

To differentiate these, treat y as a function of x and apply the chain rule whenever you differentiate a term containing y.

Use the Chain Rule Calculator for explicit chain rule applications. This guide covers the implicit case where chain rule is embedded in every y-term.

Why Chain Rule Is the Core of Implicit Differentiation

When y is a function of x (even if you don’t know the explicit formula), differentiating any expression in y with respect to x requires the chain rule:

d/dx[y^n] = n·y^(n−1) · dy/dx (chain rule: outer = power, inner = y(x))

d/dx[sin(y)] = cos(y) · dy/dx (chain rule: outer = sin, inner = y(x))

d/dx[e^y] = e^y · dy/dx (chain rule: outer = e^, inner = y(x))

The dy/dx factor at the end is the inner derivative — the derivative of y with respect to x. That factor is what you solve for.

Step-by-Step Process

Step 1: Differentiate both sides of the equation with respect to x.

Step 2: Apply chain rule to every y-term — multiply by dy/dx.

Step 3: Collect all terms containing dy/dx on one side.

Step 4: Factor out dy/dx.

Step 5: Divide to isolate dy/dx.

Worked Example 1 — Circle

Find dy/dx for x² + y² = 25

Step 1 — Differentiate both sides:

d/dx[x²] + d/dx[y²] = d/dx[25]
2x + 2y·(dy/dx) = 0

(d/dx[y²] = 2y·dy/dx by chain rule: power rule with inner = y)

Step 2 — Solve for dy/dx:

2y·(dy/dx) = −2x
dy/dx = −2x / 2y = −x/y

Answer: dy/dx = −x/y

This tells the slope of the circle at any point (x, y). At (3, 4): dy/dx = −3/4.

Worked Example 2 — Trig Equation

Find dy/dx for sin(x + y) = y²

Step 1 — Differentiate both sides:

Left side: d/dx[sin(x + y)] = cos(x + y) · d/dx[x + y] = cos(x + y) · (1 + dy/dx)

(Chain rule: outer = sin, inner = x + y, and d/dx[x + y] = 1 + dy/dx)

Right side: d/dx[y²] = 2y · dy/dx

Combined equation:

cos(x + y) · (1 + dy/dx) = 2y · dy/dx

Step 2 — Expand and collect dy/dx terms:

cos(x + y) + cos(x + y)·dy/dx = 2y·dy/dx
cos(x + y) = 2y·dy/dx − cos(x + y)·dy/dx
cos(x + y) = dy/dx · [2y − cos(x + y)]

Step 3 — Solve:

dy/dx = cos(x + y) / [2y − cos(x + y)]

Worked Example 3 — Mixed Exponential

Find dy/dx for e^(xy) = x + y

Step 1 — Differentiate both sides:

Left side: d/dx[e^(xy)] = e^(xy) · d/dx[xy]

For d/dx[xy], use product rule: d/dx[x·y] = 1·y + x·dy/dx = y + x·dy/dx

So: d/dx[e^(xy)] = e^(xy) · (y + x·dy/dx)

Right side: d/dx[x + y] = 1 + dy/dx

Combined:

e^(xy) · (y + x·dy/dx) = 1 + dy/dx

Step 2 — Expand:

y·e^(xy) + x·e^(xy)·dy/dx = 1 + dy/dx

Step 3 — Collect dy/dx terms:

x·e^(xy)·dy/dx − dy/dx = 1 − y·e^(xy)
dy/dx · (x·e^(xy) − 1) = 1 − y·e^(xy)

Step 4 — Solve:

dy/dx = (1 − y·e^(xy)) / (x·e^(xy) − 1)

Worked Example 4 — Finding the Tangent Line

Find the equation of the tangent line to x³ + y³ = 9 at point (1, 2)

Step 1 — Differentiate implicitly:

3x² + 3y²·(dy/dx) = 0
3y²·(dy/dx) = −3x²
dy/dx = −x²/y²

Step 2 — Evaluate at (1, 2):

dy/dx = −(1)²/(2)² = −1/4

Step 3 — Write tangent line using point-slope form:

y − 2 = −(1/4)(x − 1)
y = −x/4 + 1/4 + 2
y = −x/4 + 9/4

Tangent line: y = −x/4 + 9/4

Common Mistakes in Implicit Differentiation

Forgetting dy/dx on y-terms: d/dx[y²] ≠ 2y. It equals 2y·(dy/dx). Every differentiation of a y-term produces a dy/dx factor — this is the chain rule.

Applying power rule to x and y terms the same way: d/dx[x²] = 2x (no extra factor). d/dx[y²] = 2y·dy/dx (dy/dx required). The difference is that x is the variable, y is a function of x.

Stopping before isolating dy/dx: After expanding, you must factor out dy/dx and divide to get a single dy/dx = expression. Many students leave the answer with dy/dx on both sides.

When to Use Implicit vs Explicit Differentiation

For chain rule applied to fully explicit composite functions, see How to Apply the Chain Rule Step by Step. For the complete pattern reference used in these examples, see the Chain Rule Patterns Reference.