Implicit Differentiation

Implicit differentiation finds $dy/dx$ even when an equation never isolates $y$. Instead of solving for $y$ first, you differentiate both sides with respect to $x$ and treat $y$ as a function of $x$.

When To Use This Method

Reach for implicit differentiation when the curve is given by a relation rather than a clean $y = f(x)$. Some curves are easy to write with one equation but awkward as a single formula. A circle is the standard case:

This represents the whole circle at once. Solving for $y$ would split it into a top branch and a bottom branch, but implicit differentiation reads the slope straight from the original relation. Concretely, it is the right tool when:

1. A curve is given by a relation such as a circle, ellipse, or level curve.
2. Solving explicitly for $y$ would be messy or split the curve into separate cases.
3. You need the slope of a tangent line at a point.
4. A related-rates problem links changing variables before differentiating with respect to time.

The general setup is a relation $F(x,y) = 0$. If it defines $y$ as a differentiable function of $x$ near the point you care about, you can differentiate the whole equation with respect to $x$ and solve for $dy/dx$.

The Procedure, Step By Step

1. Differentiate every term with respect to $x$.
2. Treat $y$ as changing with $x$, so each derivative of a $y$-term picks up a chain-rule factor.
3. Solve the resulting equation for $dy/dx$.

Step 2 is the one students usually miss. For example,

not just $2y$.

The Whole Procedure On One Example

Find $dy/dx$ for

Differentiate both sides with respect to $x$ (Step 1), attaching the chain-rule factor to the $y$-term (Step 2):

Now solve for $dy/dx$ (Step 3):

This formula works at points where $y \ne 0$. If $y = 0$, dividing by $y$ is invalid, and on this circle those points correspond to vertical tangents. At the point $(3,4)$,

so the tangent line slopes downward there.

Where Each Step Trips People Up

The chain rule enters whenever you differentiate a term containing $y$, because $y$ depends on $x$:

and

A quick self-check: if you finish differentiating an expression that contained $y$ and see no $dy/dx$ term, go back, because you dropped a chain-rule factor. The recurring stumbles are:

1. Differentiating $y^2$ as $2y$ instead of $2y \frac{dy}{dx}$.
2. Forgetting that a mixed term such as $xy$ needs the product rule.
3. Dividing by an expression to isolate $dy/dx$ without checking whether it could be $0$.
4. Assuming one derivative formula holds globally, even when the relation has multiple branches.

A Harder Run-Through

Try

Differentiate both sides and solve for $dy/dx$. This one exercises every step: the term $xy$ forces the product rule, while $y^2$ still produces a chain-rule factor. Working the product-rule term and the chain-rule term separately first, then combining, is the cleanest way to see where each rule lands.