Implicit Differentiation

You’ll be able to

Differentiate both sides of an equation with respect to x, treating y as a function of x
Solve the resulting equation algebraically for dy/dx
Find the slope of a tangent line to an implicitly defined curve

When you cannot solve for y

Many curves — a circle x² + y² = 25, for instance — cannot be written cleanly as y = f(x). Implicit differentiation handles them without solving for y first. The key idea is to treat y as an unknown function of x and differentiate both sides of the equation with respect to x, applying the chain rule every time y appears.

Every y picks up a dy/dx

Because y is a function of x, differentiating any expression in y triggers the chain rule and attaches a factor of dy/dx. So d/dx[y²] = 2y·(dy/dx), and d/dx[y³] = 3y²·(dy/dx). Terms in x alone differentiate normally. Products that mix x and y (like x·y) still need the product rule. Forgetting the dy/dx on a y-term is the defining implicit-differentiation error.

Then solve for dy/dx

After differentiating, the equation contains dy/dx terms scattered around. Collect every dy/dx term on one side, factor out dy/dx, and divide to isolate it. The answer usually depends on both x and y, which is expected — an implicit curve can have different slopes at different points sharing the same x-value.

Chain rule on y-terms

d/dx[yⁿ] = n·yⁿ⁻¹ · (dy/dx)

Treat y as a function of x. Any term containing y produces a dy/dx factor; terms in x alone do not.

Worked example

Find dy/dx for the circle x² + y² = 25, then the slope of the tangent at the point (3, 4).

1.Differentiate both sides with respect to x: d/dx[x²] + d/dx[y²] = d/dx[25].
2.That gives 2x + 2y·(dy/dx) = 0.
3.Solve for dy/dx: 2y·(dy/dx) = −2x, so dy/dx = −x/y.
4.Substitute the point (3, 4): dy/dx = −3/4.

Answer: dy/dx = −x/y, and at (3, 4) the tangent slope is −3/4. (This matches the geometric fact that a tangent is perpendicular to the radius.)

Watch out

Differentiating y² as just 2y is the number-one mistake. It must be 2y·(dy/dx) because y is a function of x. Every y-term earns its own dy/dx factor from the chain rule.

Checkpoint

What is d/dx[y³] when y is a function of x?

Checkpoint

For the curve xy = 6, differentiating implicitly gives which equation?

On the exam

To find a tangent slope on an implicit curve you need both coordinates, because dy/dx typically depends on x and y. Differentiate, solve for dy/dx, then substitute the full point — not just the x-value.

Answer the 2 checkpoints as you read.